UC-NRLF 


35    fibl 


GIFT  OF 


SOLID   GEOMETRY 


SOLID  GEOMETRY 


By 
MABEL   SYKES 

Instructor  in  Mathematics,  Bowen  High  School,  Chicago 
Author  of  "A  Source- Book  of  Problems  for  Geometry" 
and  (with  Clarence  E.  Comstock)  of  "Plane  Geometry" 


and 


CLARENCE   E.   COMSTOCK 

Professor  of  Mathematics,  Bradley  Polytechnic  Institute 
Author  (with  Mabel  Sykes)  of ''Beginners'  Algebra" 


RAND   M9NALLY  &   COMPANY 

CHICAGO  NEW  YORK 


J  of      '  '   . 


Copyright,  1922,  by 
RAND  MCNALLY  &  COMPANY 


/ 

r 


Made  in  U.  S.  A. 


THE   CONTENTS 

PAGE 

The  Preface vii 

CHAPTER  I.     Lines  and  Planes 

Introductory 1 

Parallel  Lines  and  Planes 6 

Perpendicular  Lines  and  Planes 12 

Equal  Angles  in  Space 17 

Perpendiculars  and  Parallels 18 

Dihedral  Angles .  21 

Perpendicular  Planes 23 

Proportional  Segments 29 

Loci  in  Space 29 

Projection 35 

Polyhedral  Angles 38 

Summary  and  Supplementary  Exercises 42 

CHAPTER  II.     Properties  of  Polyhedrons,  Cylinders,  and  Cones 

Solids  in  General 48 

Some  Elementary  Surfaces 48 

Prisms 51 

Cylinders 55 

Pyramids 60 

Cones 64 

Regular  Polyhedrons 69 

Supplementary  Exercises 72 

CHAPTER  III.    The  Sphere 

Introductory 76 

Tangents  to  Spheres 77 

Determination  of  Spheres 79 

Circles  of  Spheres 81 

Spherical  Angles 86 

Spherical  Triangles 87 

Supplementary  Exercises •      •      .102 


vi  THE   CONTENTS 

CHAPTER  IV.     Areas  and  Volumes 

PAGE 

Areas  of  Polyhedrons 106 

Volumes  of  Polyhedrons 109 

The  Measurement  of  Round  Bodies  in  General 125 

The  Measurement  of  the  Cylinder 125 

The  Measurement  of  the  Cone 129 

The  Measurement  of  Frustums  of  Cones 132 

General  Formula 134 

Volumes  by  Cavalieri's  Theorem 138 

Spherical  Measurements 142 

Summary  and  Supplementary  Exercises 156 

CHAPTER  V.     Similarity  and  Symmetry 

Similarity 162 

Symmetry     .      .      .      .      '. 172 

Miscellaneous  Exercises  173 


Notes  on  Arithmetic  and  Algebra         174 

Tables 180 

Outline  of  Plane  Geometry 185 

References  and  Topics  for  Mathematics  Clubs 202 

The  Index  209 


THE   PREFACE 

The  Solid  Geometry  is  prepared  for  the  same  purpose  and 
with  the  same  general  features  as  is  the  Plane  Geometry. 
In  both  books  the  two  main  characteristics  are  analysis  and 
emphasis. 

One  of  the  great  objectives  in  education  is  to  train  young 
people  to  attack  difficulties  through  an  analysis  of  the 
problems  presented.  It  is  because  of  this  fact  that  both  the 
Plane  and  the  Solid  Geometry  are  prepared  as  suggestive 
method  texts  with  the  suggestions  in  the  form  of  a  logical 
analysis. 

Moreover,  if  the  mind  is  to  retain  any  lasting  impression 
of  the  work  covered,  distinctions  in  emphasis  are  necessary. 
The  material  presented  in  both  the  Plane  and  the  Solid 
Geometry  has  been  arranged  with  this  fact  in  mind.  Atten- 
tion is  called  particularly  to  chapters  ii,  iv,  and  v.  Chap- 
ter ii  discusses  the  nature  and  properties  of  the  various  sur- 
faces and  solids  ordinarily  studied  in  solid  geometry  with 
the  exception  of  the  sphere,  which  is  studied  in  chapter  iii. 
All  areas  and  volumes  are  considered  in  chapter  iv.  Simi- 
larity is  considered  in  chapter  v. 

There  are  several  advantages  gained  from  this  arrangement. 
It  enables  the  pupil  to  take  up  the  study  of  areas  and 
volumes  with  a  clear  idea  of  the  solids  considered.  It  makes 
possible  a  more  logical  arrangement  of  material.  Cylinders 
are  compared  with  prisms  and  cones  with  pyramids  when 
the  properties  of  these  solids  are  studied;  but  as  the  volume 
of  the  pyramid  is  obtained  from  that  of  the  prism,  a  different 
order  is  used  when  volumes  are  studied.  Moreover,  -the 
theorems  concerning  areas  and  volumes  may  be  worked  into 
a  logical  whole  when  considered  together,  which  is  not  possi- 
ble in  the  traditional  arrangement.  For  example,  Theorems 
121,  and  122  serve  not  only  as  a  necessary  preparation  for 

vii 


viii  THE    PREFACE 

the  measurement  of  the  sphere,  but  also  as  a  fitting  climax 
to  the  work  on  cylinders  and  cones. 

Both  the  Plane  and  the  Solid  Geometry  are  written  with  the 
firm  conviction  that  if  geometry  is  taught  by  analysis  and 
if  at  the  same  time  proper  distinctions  in  emphasis  are  made, 
pupils  will  reach  the  end  of  their  course  with  more  real 
education  and  with  a  much  clearer  and  more  lasting  impres- 
sion of  the  meaning  of  the  great  concepts  of  geometry  than 
can  possibly  be  the  case  under  traditional  methods. 

Attention  is  called  also  to  certain  minor  features: 

The  approach  to  the  early  theorems  through  the  intro- 
ductory material  is  natural  and  easy. 

As  the  first  difficulty,  perhaps  the  only  real  difficulty,  in 
solid  geometry,  is  the  inability  of  pupils  to  visualize  figures 
in  space,  the  use  of  models  made  by  pupils  is  strongly  recom- 
mended. These  models  should  be  used  for  demonstration 
work  and  should  precede  the  use  of  blackboard  figures  until 
the  pupil  is  clearly  out  of  "flatland."  If  possible,  spherical 
triangles  should  be  studied  from  a  slated  globe. 

The  treatment  of  loci  in  §§  41-47  and  the  treatment  of 
similarity  in  chapter  v  deserve  attention. 

The  formal  study  of  the  theory  of  limits  is  omitted.  The 
treatment  of  areas  and  volumes  of  round  bodies  by  this 
method  is  given  in  outline  only.  It  is  intended  merely  to 
make  the  results  appear  reasonable  to  the  pupil. 

By  proper  choice  of  material  the  study  of  volumes  may  be 
made  to  depend  upon  Cavalieri's  theorem.  See  §§  173-179 
and  §  154. 

The  ''Notes  on  Arithmetic  and  Algebra"  and  the  ''Outline 
of  Plane  Geometry"  given  will  be  found  convenient  for 
reference.  The  "Topics  and  References  for  Mathemalic 
Clubs"  are  intended  as  suggestions  only. 

M.  S. 

CHICAGO,  ILLINOIS  C.  E.  C. 

March,  1922 


SOLID    GEOMETRY 

CHAPTER   I 

LINES  AND  PLANES 

INTRODUCTORY 

SUBJECT   MATTER   OF   GEOMETRY 

1.  In  plane  geometry  our  study  was  confined  to  figures 
that  could  be  drawn  on  a  plane  and  drawn  with  ruler  and 
compass  only.     Such  figures  are  constructed  of  points  and 
lines. 

In  solid  geometry  we  extend  our  study  to  figures  in 
space.  These  figures  are  constructed  of  points,  lines,  sur- 
faces, and  solids. 

2.  The  space  occupied  by  a  ball  is  a  geometrical  solid. 
The  outside  of  the  ball  is  a  surface;  it  is  the  boundary  of  the 
solid   and   separates   the   space  within 

from  the  space  without  (Fig.  1). 

In  general  we  may  say  that — 

The  space  occupied  by  any  object  is 
called  a  geometrical  solid. 

The  boundaries  of  a  solid  are  called 
surfaces. 

The  surfaces  of  a  solid  separate  the  FIG  i 

solid  from  the  remainder  of  space. 

We  may  define  a  surface  as  any  boundary  between 
two  parts  of  space. 

The  space  occupied  by  a  tomato  can  is  a 
geometrical  solid.  The  surface  of  the  can 
separates  the  space  within  from  the  space  with- 
out. We  may  consider  the  total  surface  of  the 
can  as  composed  of  three  parts  separated  by 
the  edges  of  the  can  (Fig.  2).  FIG.  2 

1 


SOLID   GEOMETRY 


The  edges  of  a  solid  are  called  lines.  We  may  consider 
the  edges  of  a  solid  as  separating  the  parts  of  its  surface. 

We  may  define  a  line  as  the  boundary  between  two  parts 
of  any  surface. 

The  equator  is  a  line  separating  the  northern  and  south- 
ern hemispheres,  two  parts  of  the  surface  of  the  earth. 

The  space  occupied  by  a  rectangular  block  is  a  geomet- 
rical solid.     The  lateral  surface 
of  the  block  (Fig.  3)  consists  of 
four  parts  which  we  may  think 
of  as  separated  by  the  lateral         j 
edges,  or  lines.    The  edge  about 

"F*Tr*     ^ 

the  upper  base  may  be  consid- 
ered as  composed  of  four  parts  separated  by  the  corners,  or 
points. 

The  corners  of  a  solid  are  called  points.  We  may  con- 
sider the  corners  of  a  solid  as  separating  the  parts  of  its  edges. 

We  may  define  a  point  as  the  boundary  between  two 
parts  of  any  line. 

A  point  on  a  straight  line  divides  the  line  into  two  parts 
called  rays. 

DETERMINATION   OF  PLANES 

3.  The  following  assumptions  concerning  planes  are 
important : 

As.  1.  A  straight  line  joining  any  two  points  in  a  plane 
lies  wholly  in  the  plane. 

This  is  the  fundamental  characteristic  of  planes. 

Exercise.  Is  it  possible  to  find  two  points  on  the  curved  sur- 
face of  a  tomato  can,  such  that  the  line  joining  them  lies  wholly  on 
the  surface  ?  Why  is  this  surface  not  a  plane  ? 

As.  2.  A  plane  is  unlimited  in  extent. 
As.  3.  Through  a  given  straight  line  an  unlimited  number 
of  planes  may  be  passed. 


LINES  AND   PLANES  3 

If  a  number  of  planes  pass  through  a  given  line,  they 
are  said  to  form  a  pencil  of  planes,  and  the  line  is  said  to 
be  the  axis  of  the  pencil. 

In  Fig.  4,  AB  is  said  to  be  the' axis  of  the  pencil  of  planes. 

As.  4.  If  a  plane  passes  through  a  given  straight  line, 
it  may  be  revolved  so  as  to  contain  any 
given  point  in  space.  C/^x^xV 

As.  5.  Through  a  given  straight  line     L — /^fC.    ) — 
and  a  given  point  without  the  line  only  ^ 

one  plane  can  be  passed. 

COR.  I.  A  plane  is  located  definitely  if  it  contains  three 
given  points  not  in  the  same  straight  line. 

Suggestion.  Show  that  this  is  equivalent  to  locating  the  plane  by 
a  straight  line  and  a  point  without  the  line. 

COR.  II.  A  plane  is  located  definitely  by  two  given 
intersecting  straight  lines. 

Suggestion.  Show  that  this  is  equivalent  to  locating  the  plane  by 
one  of  the  lines  and  any  point  in  the  other.  See  As.  1. 

COR.  III.  A  plane  is  located  definitely  by  two  parallel 
straight  lines. 

Suggestion.  By  definition  two  parallel  lines  lie  in  the  same  plane. 
Only  one  plane  can  contain  the  two  parallel  lines.  If  there  could  be 
two  planes  each  containing  the  two  parallels,  there  could  be  two 
planes  each  containing  a  line  and  the  same  point  without  the  line. 
This  is  impossible. 

RELATIVE   POSITIONS    OF   TWO    PLANES 

4.  Two  planes  either  intersect  or  do  not  intersect. 
Two  planes  that  do  not  intersect  are  said  to  be  parallel. 
This  definition  is  the  fundamental  test  for  parallel  planes. 
If  two  planes  intersect,  the  intersection  of  the  two  planes 
is  the  locus  of  the  points  common  to  the  two  planes. 

As.  6.  Two  intersecting  planes  have  at  least  two  common 
points. 


4  SOLID   GEOMETRY 

5.  THEOREM   1.    The  intersection  of  two  planes  is  a 
straight  line. 

Analysis:  To  prove  that  the  intersection  is  a  straight 
line,  join  two  points  in  the  intersection  by  a  straight  line 
and  prove  that 

I.  Every  point  in  this  line  lies  in  each  plane. 

II.  No  point  outside  this  line  lies  in  both  planes. 

Theorem  1  may  be  stated t thus:  The  intersection  of  two 
given  planes  determines  a  straight  line. 

RELATIVE   POSITIONS    OF   A   STRAIGHT   LINE 
AND   A   PLANE 

6.  If  a  straight  line  lies  in  a  plane,  the  plane  is  said  to 
contain  the  line. 

If  the  plane  does  not  contain  the  line,  the  line  may 
intersect  the  plane,  or  it  may  not  intersect  the  plane. 

If  the  line  and  the  plane  do  not  intersect,  they  are  said 
to  be  parallel.  This  definition  is  the  fundamental  test  for 
lines  parallel  to  planes. 

When  a  given  straight  line  intersects  a  given  plane,  one 
point  and  only  one  point  is  determined;  or 

As.  7.  A  straight  line  can  intersect  a  plane  in  but  one 
point. 

RELATIVE   POSITIONS    OF   STRAIGHT   LINES   IN    SPACE 

7.  Two  straight  lines  in  a  plane  either  intersect  or  are 
parallel.     Two  straight  lines  in  space  either  intersect  or  are 
parallel  or  skew. 

Two  straight  lines  are  said  to  be  parallel  if  they  are  in  the 
same  plane  and  do  not  meet. 

Two  straight  lines  in  space  are  said  to  be  skew  if  they 
neither  intersect  nor  are  parallel. 

Exercise.  Hold  two  pencils  to  represent  two  skew  lines.  Point 
out  two  skew  lines  in  the.  room  in  which  you  are  sitting. 


LINES  AND   PLANES  5 

EXERCISES  INVOLVING  THE  DETERMINATION 
OF  LINES  AND   PLANES 

8.  1.  Will  an  object  mounted  on  three  legs  always  stand  firm? 
Why?  Will  one  mounted  on  four  legs  always  stand  firm?  Why? 

2.  What  practical  uses  are  made  of  the  fact  referred  to  in  the 
previous  exercise? 

3.  Will  three  concurrent  straight  lines  always  lie  in  the  same 
plane?     Illustrate  your  answer.     How  many  planes  may  be  deter- 
mined by  three  concurrent  straight  lines? 

Tell  how  many  planes  would  be  determined  in  each  of  the 
following  cases  (Ex.  4-10)  and  discuss  all  possibilities.  Illustrate 
your  answers  with  pencils  or  toothpicks  and  give  reasons. 

4.  Two  intersecting  straight  lines  and  a  point  not  in  the  plane 
of  the  lines. 

5.  Four  concurrent  straight  lines. 

6.  Three  concurrent  straight  lines  and  one  point. 

7.  Four  points  not  all  in  the  same  plane.     How  many  lines 
are  determined  in  this  case? 

8.  Five  points  no  four  of  which  lie  in  the  same  plane.     How 
many  lines  are  determined  in  this  case? 

9.  Three  parallel  lines  that  are  not  all  in  the  same  plane. 

10.  Four  parallel  lines  no  three  of  which  lie  in  the  same  plane. 

11.  Must  three  intersecting  straight  lines  lie  in  the  same  plane? 
Why? 

12.  Hold  three  pencils  to  illustrate  the  various  relative  positions 
of  three  straight  lines  in  space.     How  many  planes  are  determined 
in  each  case? 

13.  Answer  Ex.  12  for  four  straight  lines. 

14.  What  is  a  skew  quadrilateral?     Can  a  skew  quadrilateral 
be  a  parallelogram?     How  many  planes  are  determined  by  the 
sides  of  a  skew  quadrilateral? 

15.  Hold  two  cards  to  illustrate  the  possible  relative  positions 
of  two  planes  in  space.     How  many  lines  are  determined  in  each 
case? 

16.  Answer  Ex.  15  for  three  planes. 


6 


SOLID   GEOMETRY 


17.  Can  skew  straight  lines  be  parallel  to  the  same  plane? 
Illustrate  your  answer  with  two  pencils  and  a  card. 

18.  What  are  the  least  number  of  planes  that  can  completely 
inclose  a  space?    How  many  lines  are  determined  in  this  case? 
Illustrate  your  answer. 

19.  If  five  planes  intersect  so  as  completely  to  inclose  a  space, 
how  many  lines  are  determined? 


PARALLEL  LINES   AND   PLANES 

TEST  FOR  LINES   PARALLEL   TO   PLANES 

9.  THEOREM  2.    If  a  plane  contains  one  and  only  one 
of  two  parallel  straight  lines,  it  is  parallel  to  the  other. 


FIG.  5 

Hypothesis:     Lines  a  and  b  are  parallel  and  plane  M  con- 
tains line  a  but  not  line  b. 

Conclusion:     Line  b  is  parallel  to  plane  M. 
Analysis: 
I.  To  prove  line  b  \\  plane  M,  prove  that  line  b  cannot 

meet  plane  M. 
II.   /.  show  that  if  line  b  met  plane  M,  it  would  also  meet 

line  a. 
Proof: 

STATEMENTS 

I.  a.  Lines  a  and  b  lie  in  the 

same  plane. 

b.  Line  a  is  the  intersec- 
tion of  plane  M  and 
plane  ab. 


REASONS 

I.  a.  Two  parallel  lines 
determine  a  plane. 
b.  Line  a  lies  in  both 
planes. 


LINES  AND   PLANES 


c.  If  line  b  meets  plane  M, 
it  will  meet  line  a. 

d.  But  line  b  cannot  meet 
line  a. 

II.    .'.  line  b  cannot  meet  plane 
M  and  is  II  it. 


c.  Line  a  is  the  inter- 
section   of    plane 
M  and  plane  ab. 
d.  Why? 

II.  The  supposition  that 
line  b  would  meet 
plane  M  is  elimi- 
nated. 


Ex.  1.  Show  that  one  plane  and  only  one  may  be  passed 
through  one  of  two  skew  lines  and  parallel  to  the  other. 

Suggestion.  Let  a  and  b  represent  the  two  given  skew  lines. 
Through  any  point  in  a  draw  b'  parallel  to  b.  Lines  a  and  b'  determine 
a  plane  parallel  to  b.  Why? 

Ex.  2.  Show  that  one  plane  and  only  one  may  be  passed 
through  a  given  point  and  parallel  to  two  given  skew  lines  in 
space. 

Suggestion.  Let  a  and  b  represent  the  two  given  skew  lines  and 
0  the  given  point.  Draw  through  0,  a'  \\  a  and  b'  \\  b.  Lines  a'  and 
b'  determine  a  plane  parallel  to  line  a  and  to  line  b.  Why? 

TESTS  FOR  PARALLEL  LINES 

10.  THEOREM  3.  If  a  straight  line  is  parallel  to  a  plane, 
it  is  parallel  to  the  intersection  of  the  given  plane  with  any 
intersecting  plane  containing  the  line. 


N 


FIG.  6 


Analysis:     To  prove  a  \\  b,  prove  that  they  are  in  the  same 
plane  and  cannot  meet. 


8  SOLID  GEOMETRY 

COR.  If  a  straight  line  and  a  plane  are  parallel,  a  straight 
line  through  any  point  of  the  given  plane  parallel  to  the 
given  line  lies  wholly  in  the  given  plane. 

Analysis: 
I.  To  prove  b  lies  in  plane  M,  prove  that  b  coincides 

with  a  line  that  does  lie          a 

in  plane  M. 
II.    .*.   pass  a  plane  through  a 

and  point  P,  intersecting 

M  in  line  x,  and  prove 

that  x  coincides  with  b. 
III.  To  prove  that  x  coincides  FlG-  7 

with  b,  show  that  x  and  b  are  both  parallel  to  a 

through  P  (Fig.  7). 

11.  THEOREM  4.  If  two  parallel  planes  are  cut  by  a 
third  plane,  the  intersections  are  parallel. 


FIG.  8 


Hypothesis:  M  and  N  are  two  parallel  planes  cut  by 
plane  P  in  lines  a  and  b  respectively. 

Conclusion:     Lines  a  and  b  are  parallel. 

Analysis:  To  prove  lines  a  and  b  parallel,  prove  that 
they  are  in  the  same  plane  and  do  not  meet. 

Exercise.     Illustrate  Ths.  1-4  by  planes  from  the  room  in  which 
you  are  sitting  and  by  pencils  and  pieces  of  cardboard. 


LINES  AND  PLANES  9 

V- 

TEST  FOR  PARALLEL  PLANES 

12.  THEOREM  5.  If  two  intersecting  straight  lines  are 
parallel  respectively  to  two  other  intersecting  straight  lines, 
the  plane  of  the  first  pair  is  parallel  to  the  plane  of  the  second 
pair. 


M 


FIG.  9 

Hypothesis:    M  and  N  are  two  planes  with  lines  a  and  b 
in  plane  M  parallel  respectively  to  lines  a'  and  br  in  plane  N. 
Conclusion:    Planes  M  and  N  are  parallel. 
Analysis: 

I.  Show  that  plane  M  cannot  meet  plane  N. 
II.   .*.  show  that  if  plane  M  met  plane  N  in  any  line, 
this  line  would  be  parallel  to  each  of  the  intersect- 
ing lines  a  and  6  (Th.  3). 

III.    .*.  prove  that  a  and  b  are  each  parallel  to  plane  N. 
Proof: 

STATEMENTS  REASONS 

I.  a.  Line  a  ||  line  a'. 

b.  Line  a  ||  plane  N.  Th.  2 

c.  Similarly  line  b  \\  plane  N. 

II.  a.  Suppose  M  and  N  meet  in  some 
line.     Call  this  intersection  line  p. 

b.  .'.  a  would  be  ||  p.  Th.  3 

c.  Similarly  b  would  be  ||  p. 

III.  But  a  and  b  cannot  both  be  ||  same 

line.  Why? 

IV.  /.  M  cannot  meet  N  and  is  ||  N. 

The  pupil  should  quote  in  full  all  theorems  referred  to  by  number. 
2 


SOLID   GEOMETRY 


TEST  FOR  PARALLEL  LINES 

13.  THEOREM  6.    Two  straight  lines  parallel  to  a  third 
straight  line  are  parallel  to  each  other. 


FIG.  10 

Hypothesis:     Lines  a  and  b  are  each  parallel  to  c. 
Conclusion:     Line  a  is  parallel  to  line  b. 
Analysis  and  construction: 

I.  To  prove  a  ||  b,  prove  that  b  coincides  with  a  line 
that  is  ||  a. 

II.  Let  M  be  the  plane  of  b  and  c,  and  N  be  the  plane 
of  a  and  X,  any  point  in  b.  Let  N  intersect  M  in 
line  d.  Prove  that  d  is  ||  a  and  that  d  coincides 
with  b. 

III.  To  prove  that  d  and  b  coincide,  prove  that  they  are 
both  parallel  to  c  through  X. 

Proof: 

STATEMENTS 

I.  a.  Line  c  \\  N. 

b.  Line  d  \\  line  c,  through  X. 

c.  Line  b  \\  line  c,  through  X. 

d.  .'.  lines  d  and  b  coincide. 


REASONS 

I.  a.  Th.  2 
6.  Th.  3 

c.  Hypothesis 

d.  As.   30,  Plane 
Geometry 

II.  Ths.  2  and  3 

Why? 


II.  but  line  d  \\  a. 
.'.  line  b  \\  a. 

NOTE.     Th.  6  is  proved  here  for  the  sake  of  the  exercises  in  §14. 
Another  proof  is  given  on  page  19. 


LINES  AND  PLANES  11 

EXERCISES   INVOLVING  PARALLELS 
14.  To  prove  two  planes  parallel,  prove  that 

1.  They  cannot  meet,  or 

2.  Two  intersecting  lines  in  one  are  parallel  respec- 
tively to  two  intersecting  lines  in  the  other. 

To  prove  a  line  and  a  plane  parallel,  prove  that 

1.  They  cannot  meet,  or 

2.  The  plane  contains  a  line  parallel  to  the  given  line. 

To  prove  two  lines  parallel,  prove  that 

1.  They  are  in  the  same  plane  and  do  not  meet,  or 

2.  One  is  parallel  to  a  given  plane  and  the  other  is 
the  intersection  of  the  given  plane  and  any  plane 
containing  the  line,  or 

3.  They  are  the  intersection  of  two  parallel  planes 
cut  by  a  third,  or 

4.  They  are  parallel  to  the  same  line. 

1.  Are  two  lines  parallel  to  the  same  plane  parallel  to  each 
other?    Hold  a  card  and  two  pencils  so  as  to  illustrate  your  answer. 

2.  Are  two  planes  parallel  to  the  same  line  parallel  to  each 
other?    Illustrate  your  answer. 

3.  Through  a  given  point  how  many  lines  can  be  drawn  parallel 
to  a  given  plane?    Why? 

4.  Parallel  segments  between  parallel  planes  are  equal  and  cut 
off  on  the  given  planes  segments  that  are  equal  and  parallel. 

5.  If  a  line  is  parallel  to  a  given  plane,  what  is  its  relation  to 
the  lines  of  the  plane?     Illustrate  your  answer.    To  what  lines  of 
the  plane  is  it  parallel? 

6.  If  a  straight  line  and  a  plane  are  parallel,  a  pencil  of  planes 
through  the  given  line  intersects  the  given  plane  in  parallel  lines. 

7.  If  a  line  is  parallel  to  one  of  two  parallel  planes,  it  is  parallel 
to  the  other  or  lies  in  the  other. 

Suggestion.     Pass  a  plane  through  the  given  line  so  as  to  cut  the 
two  parallel  planes. 

8.  If  one  of  two  parallel  lines  is  parallel  to  a  given  plane,  the 
other  is  also  parallel  to  the  plane  or  lies  in  the  plane. 


12  SOLID   GEOMETRY 

PERPENDICULAR  LINES   AND   PLANES 
PRELIMINARY  THEOREMS 

15.  THEOREM  7.    In  space  but  one  line  can  be  drawn 
perpendicular  to  a  given  line  from  a  given  point  without 
the  line. 

THEOREM  8.  In  space  any  number  of  lines  can  be  drawn 
perpendicular  to  a  given  line  through  a  given  point  on  the 
line. 

The  proofs  to  Ths.  7  and  8  are  left  to  the  pupil. 

FUNDAMENTAL   TEST   FOR    PERPENDICULARS    TO    PLANES 

16.  THEOREM  9.    If  a  line  is  perpendicular  to  each  of  two 
lines  at  their  intersection,  it  is  perpendicular  to  all  lines  that 
are  in  their  plane  and  are  drawn  through  their  intersection. 


>  \ \  \ 

~osx«~-;r^:r/;^  ~a     \ 


FIG.  11 

Hypothesis:     Line  AO  _L  lines  a  and  b  at  their  intersection 
0.     M  is  the  plane  of  a  and  b. 

Conclusion:     AO  _L  all  lines  in  M  through  0. 
Analysis  and  construction: 

I.  To  prove  AO  _L  all  lines  in  M  through  0,  draw 

c  any  line  in  M  through  0  and  prove  AO  J_  c. 

II.  Draw  any  line  in  M  intersecting   a,   b,   and  c  in 

X,  Y,  and  Z,  respectively.     Extend  AO,  making 

AO  =  OB.    Join  AX,  AZ,  AY.  BX,  BZ,  and  BY. 


LINES  AND  PLANES 


13 


III.  To  prove  AO  _L  c,  prove  ZA  =  ZB  (Plane  Geometry, 
Th.  86,  Cor.). 


IV.    /.  prove   AAXZ^ABXZ. 

V.    /.        " 


=  Z.BXZ. 


VI.  To  prove  £AXZ  =  Z.BXZ,  prove  &AXY  & 
ABXY. 


VII. 


prove 


=  BXa.ndAY  = 


In  the  same  way  AO  may  be  proved  perpendicular  to  all 
lines  in  M  through  0. 

Exercise.  Make  a  model  of  toothpicks,  string,  and  cardboard 
to  illustrate  Fig.  1  1  and  give  the  analysis  above  from  your  model. 

The  intersection  of  a  line  and  a  plane  is  called  the  foot 
of  the  line. 

A  line  that  is  perpendicular  to  all  lines  in  a  plane  passing 
through  its  foot  is  said  to  be  perpendicular  to  the  plane. 

Theorem  9  may  be  stated  :  If  a  line  is  perpendicular  to 
each  of  two  lines  at  their  point  of  intersection,  it  is  per- 
pendicular to  their  plane. 

COR.  If  a  line  is  perpendicular  to  a  plane,  it  is  perpen- 
dicular to  all  lines  in  the  plane  passing  through  its  foot. 

17.  THEOREM  10.  If  from  a  point  in  a 
perpendicular  to  a  plane  equal  oblique  seg- 
ments are  drawn  to  the  plane,  they  cut  off 
equal  distances  on  the  plane  from  the  foot 
of  the  perpendicular  and  conversely  (Fig.  12). 

Exercise.  If  from  a  point  in  a 
perpendicular  to  a  plane  unequal 
oblique  segments  are  drawn  to  the 
plane,  the  longer  segment  cuts 
off  the  greater  distance  from 
the  foot  of  the  perpendicular 
and  conversely  (Fig.  13). 


FIG.  12 


FIG.  13 


14 


SOLID   GEOMETRY 
DETERMINATION    OF   PLANES 


18.  THEOREM  11.    Not  more  than  one  plane  can  be  drawn 
containing  a  given  point  and  perpendicular  to  a  given  line. 

Case  A.     When  the  given  point  is  on  the  given  line. 

A 


B 

FIG.  15 


FIG.  14 
Analysis  and  construction  (Fig.  14) : 

I.  Suppose  that  two  planes  M  and  N  are  both  J_  AB 

at  0. 

II.  Show  that  this  supposition  would  give  two  lines  in 

the  same  plane  _L  the  same  line  at  the  same  point. 
III.    /.  pass  any  plane  through  AB  intersecting  M  and  N 

in  lines  a  and  b  respectively. 
Outline  of  proof: 

I.  a.  Line  a  J_  AB  at  0  in  plane  P. 

b.  Line  b  _L  AB  at  0  in  plane  P. 

c.  This  is  impossible.    (Plane  Geometry,  As.  7.) 
II.    /.  M  and  N  are  not  both  _L  AB  at  0. 

Case  B.     When  the  given  point  is  without  the  given  line. 
Analysis  and  construction  (Fig.  15) : 

I.  Suppose  that  two  planes  M  and  N  are  both  _L  AB 
from  0. 

II.  Show  that  this  supposition  would  give  two  lines  _L 

A  B  from  a  point  without  AB. 

III.  .*.  join  0  with  X  and  Y,  the  intersection  of  AB  with 

M  and  N  respectively.    Prove  OX  and  OY  J_  AB. 


LINES  AND   PLANES 


15 


We  will  assume  from  Th.  9  that  one  plane  can  be  drawn 
perpendicular  to  a  given  line  and  contain  a  given  point  (see 
§57,  Ex.  1  and  2). 

19.  THEOREM  12.  All  perpendiculars  to  a  given  straight 
line  at  a  given  point  lie  in  a  plane  that  is  perpendicular  to 
the  line  at  the  given  point. 

A 


Suggestion.  Prove  that  the  plane  deter- 
mined by  OX  and  OY  coincides  with  the 
plane  determined  by  O  Y  and  OZ.  .'.  prove 
planes  XOY  and  YOZ  each  J_  AB  at  O. 


DETERMINATION   OF  LINES 

20.  THEOREM  13.    Not  more  than  one  perpendicular  can 
be  drawn  to  a  given  plane  from  a  given  point. 

Case  A.     When  the  point  is  in  the  plane. 


Suggestion.  Suppose  a  and  b 
(Fig.  17)  are  two  perpendiculars 
to  M  from  O.  Let  the  plane  of 
a  and  b  intersect  M  in  line  RS. 
Show  that  there  would  be  at 
point  0  two  lines  in  the  same 
plane  perpendicular  to  line  RS. 


FIG.  17 


Case  B.     When  the  point  is  without  the  plane. 

o 


Suggestion.  How  is  line  RS  (Fig. 
18)  determined?  Show  that  there  would 
be  from  point  O  two  lines  in  the  same 
plane  perpendicular  to  the  line  RS. 


FIG.  18 


We  will  assume  that  one  line  can  be  drawn  perpendicular 
to  a  given  plane  from  a  given  point  (see  §57,  Ex.  3  and  4). 


16  SOLID  GEOMETRY 

COMPARATIVE  LENGTHS   OF   SEGMENTS 

21.  THEOREM  14.    The  perpendicular  from  a  point  to  a 
plane  is  the  shortest  distance  from  the  point  to  the  plane. 


FIG.  19 

Suggestion.  If  OA  is  not  the  shortest  distance  from  0  to  plane  M , 
suppose  OB  to  be  the  shortest  distance.  Find  the  intersection  of 
plane  AOB  and  M.  Show  that  this  contradicts  Th.  56,  Plane  Geometry, 
the  perpendicular  is  the  shortest  segment  from  a  point  to  a  straight 
line. 

EXERCISES  INVOLVING   PERPENDICULARS 

22.  1.  In  Fig.  20,  AO  is  perpendicular 
to  plane  M.  OB  is  perpendicular  to  line 
x,  any  line  in  M.  Prove  that  AB  is  per- 
pendicular to  x  and  that  x  is  perpendicular 
to  plane  AOB. 

2.  In  Fig.  20,  suppose  that  AO  is  per- 
pendicular to  plane  M  and  that  AB  is  drawn 

from  point  A  perpendicular  to  line  x.  Line  x  is  any  line  in  M. 
Prove  that  BO  is  perpendicular  to  line  x  and  that  x  is  perpendicular 
to  plane  AOB. 

3.  In  Fig.  21,  AB  is  perpendicular  to  plane  M  and  BC  is  oblique 
to  plane  M.    CD  and  CE  are  two  lines  in  plane  M  making  equal 
angles  with  CB.     Prove  that  CD  and  CE 

make  equal  angles  with  CA . 

4.  In  Fig,  21,  suppose  that  AB  is  drawn 
perpendicular  to  plane  M  and  BC  is  oblique 
to  plane  M .    If  CD  and  CE  are  in  plane  M 
and  make  equal  angles  with  A  C,  prove  that 

they  make  equal  angles  with  line  BC. 

FIG.  21 


FIG.  20 


LINES  AND   PLANES  \    17 

5.  If  CA  bisects  /.DCE  and  AB  is  perpendicular  to  the  plane 
of   /.DCE,  prove  that  any  point  in  AB  is  equally  distant  from 
CD  and  CE.     (Use  Ex.  2,  p.  16.) 

6.  If  three  segments  are  equal  and  parallel  and  are  not  all  in 
the  same  plane,  the  segments  joining  corresponding  extremities 
form  congruent  triangles.   • 

EQUAL  ANGLES  IN  SPACE 

23.  THEOREM  15.  If  two  angles  lying  in  different  planes 
have  their  sides  respectively  parallel  and  lie  on  the  same 
side  of  the  line  joining  their  vertices,  the  angles  are  equal. 


FIG.  22 

Hypothesis:  ABAC  and  B'A'C',  lying  in  planes  N  and 
M,  have  AB  \\  A'B',  AC  \\  A'C',  and  ABt  A'B',  AC,  and 
A'C'  lying  on  the  same  side  of  the  line  A  A'  which  joins 
their  vertices. 

Conclusion:     Z.BAC  =  £B'A'C'. 
Analysis  and  construction: 
I.  To   prove    /.A=Z.A't   prove   them   corresponding 

angles  of  congruent  triangles. 

II.    /.  make  AB  =  A'Br,  AC  =  A'C',  join  BC  and  B'C, 
and  prove  BC  =  B'C'. 

III.  To  prove  BC  =  B'C,  join  BB'  and  CC  and  prove 

BB'C'Ca  O. 

IV.  /.  prove  BB'  \\  and  =  07'. 

V.    /.  prove  BB'  and  CC'  each  ||  and  equal  to  AAr. 
VI.    .'.  prove  AA'B'B  and  AA'CC  ZS7. 
The  proof  is  left  to  the  pupil. 


18  SOLID   GEOMETRY 

PERPENDICULARS  AND   PARALLELS 
TEST   FOR   PERPENDICULARS   TO   PLANES 

24.  THEOREM  16.    If  one  of  two  parallel  lines  is  perpen- 
dicular to  a  plane,  the  other  is  also  perpendicular  to  the  plane. 

F 
B 


M 


D E 


FIG.  23 

Hypothesis:    AB  and  DF  are  two  parallel  straight  lines 
cutting  plane  M  at  A  and  D  respectively.     AB  J_  plane  M. 
Conclusion:     DF  JL  plane  M. 
Analysis  and  construction. 

I.  To  prove  DF  JL  M,  prove  DF  J_  to  two  lines  in  M. 
To  prove  DF  _L  one  line,  as  DE,  in  M,  draw  AC 
in  M  from  A  and  ||  DE  and  prove  that 
(1)  ZFDE=  ^BAC  and  (2)   Z.BAC  is  art.  Z. 
In  the  same  way  DF  can  be  proved  JL  any  other 
line  in  M  through  D. 


II. 


III. 


TEST   FOR   PARALLEL  LINES 

25.  THEOREM  17.    Two  lines  perpendicular  to  the  same 
plane  are  parallel  to  each  other. 


\ 


FIG.  24 

Hypothesis:   Lines  a  and  b  are  each  _L  plane  M. 
Conclusion:   a  II  b. 


LINES  AND   PLANES  19 

Analysis  and  construction: 
I.  To  prove  b  \\  a,  prove  that  b  coincides  with  a  line 

that  is  ||  a. 
II.    /.  construct  c  from  a  point  in  b  \\  a  and  prove  that 

c  coincides  with  b. 

III.  To  prove  that  c  and  b  coincide,  show  that  c  and  6 
are  both  JL  M  from  the  same  point. 

COR.    Two  straight  lines  parallel  to  a  third  are  parallel 
to  each  other. 

Suggestion.     If  lines  a  and  b  are  both  parallel  to  line  c,  pass  a  plane 
J.  line  c  and  prove  that  a  and  b  are  both  JL  this  plane. 

TEST  FOR  PERPENDICULARS  TO  PLANES 

26.  THEOREM  18.     If  a  straight  line  is  perpendicular  to 
one  of  two  parallel  planes,  it  is  perpendicular  to  the  other. 


FIG.  25 

Hypothesis:     Plane  M  ||  plane  N.     Line  h  _L  plane  N. 
Conclusion:     Line  h  JL  plane  M. 
Analysts  and  construction: 

I.  To  prove  line  h  J_  M,  prove  h  J_  two  lines  in  M. 
.'.  construct  BC  and  BD  any  two  lines  in  M  through 

B  and  prove  h  A.  BC  and  BD. 
To  prove  h  _L  BC  and  BD,  prove  h  JL  lines  that  are 

||  BC  and  £ZX 

IV.    /.  let  the  plane  of  h  and  BC  cut  N  in  line  AE  and  the 
plane  of  h  and  BD  cut  A/"  in  line  AF  and  prove 
BC  and  AF  II  ££>. 


II. 


Ill 


20  SOLID   GEOMETRY 

TEST  FOR  PARALLEL  PLANES 

27.  THEOREM  19.    Two  planes  perpendicular  to  the  same 
straight  line  are  parallel. 

Outline  of  proof: 
I.  If  the  two  planes  were  not  parallel,  they  would 

intersect. 

II.  Suppose  the  two  planes  intersect.     Let  P  represent 
some  point  in  the  intersection. 

III.  We  should  then  have  two  planes  containing  a  given 

point  and  perpendicular  to  a  given  line. 

IV.  This  is  impossible. 

V.    .'.    the  planes  are  parallel. 

MISCELLANEOUS  EXERCISES 

28.  1.  If  two  angles  lying  in  different  planes  have  their  sides 
parallel  and  lying  on  opposite  sides  of  the  line  joining  their  vertices, 
the  angles  are  equal. 

2.  When  are  two  angles  lying  in  different  planes  supplementary? 
Give  proof. 

3.  Two  planes  parallel  to  a  third  are  parallel  to  each  other. 
Analysis: 

I.  To  prove  plane  M  \\  plane  N,  prove  them  _J_  the  same  line. 
II.    .*.    construct  a  line  _L  the  third  plane  and  prove  this  line  J_ 
M  and  N. 

4.  Are  two  lines  perpendicular  to  the  same  plane  coplanar?    Are 
three  lines  perpendicular  to  the  same  plane  coplanar? 

5.  How  many  planes  are  determined  by  three  lines  perpendic- 
ular to  the  same  plane?     Show  that  each  of  these  planes  is  parallel 
to  one  of  the  given  lines. 

6.  Show  that  it  is  not  always  true  in  space  that  a  line  perpen- 
dicular to  one  of  two  parallels  is  perpendicular  to  the  other.    State 
this  theorem  so  that  it  is  always  true  in  space. 

7.  Do  you  know  any  other  theorems  that  may  be  true  on  a 
plane  that  are  not  true  in  space? 

8.  If  a  line  is  parallel  to  a  plane,  it  is  everywhere  equally  distant 
from  thejplane. 


LINES  AND   PLANES  21 

9.  Two  points  on  the  same  side  of  a  plane  and  equally  distant 
from  it  determine  a  line  parallel  to  the  given  plane. 

10.  Two  parallel  planes  are  everywhere  equally  distant. 

11.  If  lines  a  and  b  are  parallel  and  planes  M  and  N  are  per- 
pendicular respectively  to  lines  a  and  b,  prove  that  plane  M  is 
either  parallel  to  or  coincides  with  plane  N. 

12.  If  planes  M  and  N  are  parallel  and  lines  a  and  b  are  per- 
pendicular respectively  to  planes  M  and  N,  then  line  a  is  either 
parallel  to  or  coincides  with  line  b. 

13.  A  line  and  a  plane  perpendicular  to  the  same  line  are 
parallel  unless  the  line  lies  in  the  plane. 

14.  If  a  line  and  a  plane  are  parallel,  is  any  line  perpendicular 
to  the  given  line  perpendicular  also  to  the  given  plane?     Illustrate 
your  answer  by  using  pencils  and  pieces  of  cardboard. 

15.  Given  line  a  parallel  to  plane  M .   From  any  point  in  a  draw 
line  b  perpendicular  to  M.    Prove  that  b  is  perpendicular  to  a. 

16.  If  each  of  two  intersecting  planes  contains  one  of  two 
parallel  lines,  the  intersection  of  the  planes  is  parallel  to  each  of 
the  lines  and  to  their  plane. 

17.  AB,  CD,  and  EF  are  three  equal  segments  perpendicular 
to  plane  M  at  the  points  A ,  C,  and  E,  respectively.     Prove  that 
the  plane  determined  by  points  B,  D,  and  F  is  parallel  to  plane  M. 

DIHEDRAL  ANGLES 

29.  The  two  parts  into  which  a  straight  line  divides  a 
plane  are  called  half-planes.  The  line  is  called  the  edge  of 
the  half -plane. 

Two  half-planes  with  a  common  edge  form  a  dihedral 
angle.     The  common  edge  is  the  edge  of  the 
dihedral  angle.   The  half -planes  are  the  faces  of 
the  dihedral  angle. 

In  Fig.  26,  the  half-planes  ABDC  and  ABFE 
have  a  common  edge  AB.  AB  is  the  edge  and 
ABDC  and  ABFE  are  the  faces  of  the  dihedral 
angle.  The  angle  may  be  read  D-AB-E,  or,  if 
there  is  no  ambiguity,  AB. 


22 


SOLID   GEOMETRY 


MEASUREMENT  OF  DIHEDRAL  ANGLES 

30.  An  angle  formed  by  two  straight  lines,  one  in  each  face 
of  the  dihedral  angle,  perpendicular  to  the  edge  at  the  same 
point  is  called  a  plane  angle  of  the  dihedral 

angle.  In  Fig.  27,  Z.XOY  is  a  plane  angle 
of  the  dihedral  angle  D-AB-E  if  OX  is  in 
face  ABFE  and  OY  is  in  face  ABDC  and 
OX  and  OY  are  each  perpendicular  to  A B  at 
the  same  point  0. 

THEOREM   20.     All   plane    angles    of   the 
same  dihedral  angles  are  equal. 

The  proof  is  left  to  the  pupil. 

THEOREM  21.  If  a  plane  is  perpendicular  to  the  edge 
of  a  dihedral  angle,  its  intersections  with  the  faces  of  the 
dihedral  angle  form  a  plane  angle  of  the  dihedral  angle. 

31.  Two  dihedral  angles  are  said  to  be  congruent  if  they 
can  be  made  to  coincide.     Two  dihedral  angles  are  said 
to  be  equal  if  they  have  the  same  measure  number. 

The  measure  of  a  dihedral  angle  will  be  denned  as  the 
measure  of  its  plane  angle.  From  this  definition  we  have 

THEOREM  22.  If  two  dihedral  angles  are  equal,  their 
plane  angles  are  equal. 

THEOREM  23.  If  the  plane 
angles  of  two  dihedral  angles  are 
equal,  the  dihedral  angles  are  equal. 

Exercise.  Cut  and  fold  a  card  as 
shown  in  Fig.  28  so  as  to  illustrate  the 
definition  of  the  measure  of  a  dihedral 
angle. 

As.  8.  Two  dihedral  angles  that  are  congruent  are  equal, 
and,  conversely,  two  dihedral  angles  that  are  equal  are 
congruent. 


FIG.  28 


LINES  AND   PLANES  23 

RELATED  DIHEDRAL  ANGLES 

32.  Two  dihedral  angles  are  said  to  be  complementary 
or  supplementary  according  as  their  plane  angles  are  com- 
plementary or  supplementary. 

Two  dihedral  angles  are  said  to  be  adjacent  if  they  have 
a  common  edge  and  a  common  face  separating  the  angles. 

Two  dihedral  angles  are  said  to  be  vertical  if  the  faces  of 
one  are  prolongations  of  the  faces  of  the  other. 

33.  THEOREM    24.    Two    vertical    dihedral    angles    are 
equal. 

Suggestion.  In  this  theorem  and  in  the  following  exercises  pass  a 
plane  perpendicular  to  the  edge  of  one  of  the  dihedral  angles  and  show 
that  this  will  determine  the  plane  angles  of  all  the  dihedral  angles. 

Ex.  1.  If  one  plane  meets  another,  the  adjacent  dihedral 
angles  formed  are  supplementary. 

Suggestion.  Pass  a  plane  perpendicular  to  the  common  edge  and 
reduce  to  a  plane  geometry  theorem. 

Ex.  2.  If  two  parallel  planes  are  cut  by  a  third  plane,  the 
alternate  interior  dihedral  angles  are  equal. 

Ex.  3.  Prove  Ex.  2  for  the  corresponding  dihedral  angles. 

Ex.  4.  Can  you  state  any  other  cases  of  equal  or  supplementary 
dihedral  angles?  Give  proof. 

PERPENDICULAR  PLANES 

FUNDAMENTAL   TEST 

34.  If  the  plane  angle  of  a  dihedral  angle  is  a  right  angle, 
the  dihedral  angle  is  said  to  be  a  right  dihedral  angle  and 

the  faces  of  the  dihedral  angle  are  said  to  be  perpendicular 
to  each  other. 

If  two  planes  are  perpendicular  to  each  other,  the  dihedral 
angles  formed  are  equal  and  the  plane  angles  of  the  dihedral 
angles  are  right  angles. 

These  definitions  follow  at  once  from  the  definition  of  the 
measure  of  a  dihedral  angle. 


24 


SOLID   GEOMETRY 
TEST   FOR   PERPENDICULAR   PLANES 


35.  THEOREM  25.  If  a  straight  line  is  perpendicular  to 
a  given  plane,  every  plane  passed  through  the  line  is  per- 
pendicular to  the  given  plane. 


D 


FIG.  29 


II. 


III. 


Hypothesis:  Line  AO  _L  plane  M  at  point  0.  N  is  any 
plane  passed  through  line  AO  intersecting  plane  M  in  line  CD. 

Conclusion:     Plane  N  J_  plane  M. 
Analysis  and  construction: 

I.  To  prove  plane  N  _L  plane  M,  prove  dihedral  angle 
A-CD-B  a  right  dihedral  angle, 
prove  a  plane  angle  of  dihedral  angle  A-CD-B 
a  right  angle. 
,   draw  OB  in  plane  M  JL  CD  at  0  and  prove 

(1)  Z.AOB   the   plane   angle   of    dihedral   angle 
A-CD-B. 

(2)  Z.AOB  a  right  angle. 

Ex.  1.  If  a  plane  is  perpendicular  to  the  edge  of  a  dihedral 
angle,  it  is  perpendicular  to  each  of  its  faces. 

Ex.  2.  The  rays  that  form  a  plane  angle  of  a  dihedral  angle  lie 
in  a  plane  which  is  perpendicular  to  the  faces  of  the  dihedral  angle. 

Ex.  3.  If  three  or  more  planes  intersect  in  parallel  lines,  a 
plane  perpendicular  to  one  of  the  lines  is  perpendicular  to  all  of 
the  planes. 

Ex.  4.  A  given  line  is  oblique  to  a  given  plane.  Show  how  to 
pass  a  plane  through  the  given  line  so  that  it  will  be  perpendicular 
to  the  given  plane. 


LINES   AND   PLANES 
TEST   FOR   PERPENDICULARS   TO   PLANES 


25 


36.  THEOREM  26.  If  two  planes  are  perpendicular  to 
each  other,  a  line  drawn  in  one  of  them  perpendicular  to 
the  intersection  is  perpendicular  to  the  other. 


FIG.  30 


Hypothesis:     Plane  N  J_  plane  M,  intersecting  plane  M 
in  line  CD.     AO  is  a  line  in  plane  N  J_  CD  at  O. 
Conclusion:     Line  AO  J_  plane  M. 
Analysis  and  construction: 
I.  To  prove  AO  J_  plane  M,  prove  AO  J_  two  lines  in 

plane  M  through  point  O. 
II.  Since  AO  ±  CD,  construct  OB  in  plane  M  JL  CD 

at  O  and  prove  AO  J_  OJ5. 

III.  To  prove  AO  JL  OB,  prove  ZAOB  a  plane  angle  of 
dihedral  angle  A-CD-B. 

The  proof  is  left  to  the  pupil. 

Ex.  1.  If  a  line  and  a  plane  not  containing  the  line  are  per- 
pendicular to  the  same  plane,  they  are  parallel. 

Suggestion.  Line  a  and  plane  M  are  both  _L  plane  N.  Draw  a  line 
in  M  J_  the  intersection  of  M  and  N.  Prove  the  line  ||  a. 

Ex.  2.  If  a  line  is  parallel  to  a  given  plane,  any  plane  perpen- 
dicular to  the  line  is  perpendicular  also  to  the  given  plane. 

Suggestion.  Line  a  is  ||  plane  M  and  J_  plane  N.  Pass  any  plane 
through  a  intersecting  M  in  line  b.  Prove  b  _j_  N. 

Ex.  3.  A  plane  perpendicular  to  one  of  two  parallel  planes  is 
perpendicular  to  the  other. 

Suggestion.     Use  a  construction  line. 
3 


26 


SOLID  GEOMETRY 
DETERMINATION  OF  LINES 


37.  THEOREM  27.  If  two  planes  are  perpendicular,  a 
line  drawn  perpendicular  to  the  first  from  any  point  in  the 
second  lies  wholly  in  the  second. 


N      A 


j^l 


FIG.  31 


FIG.  32 


Hypothesis:  Planes  M  and  N  JL  each  other  and  intersect 
in  line  CD.  Line  AO  is  drawn  from  a  point  in  plane  N  _L 
plane  M. 

Conclusion:     AO  lies  wholly  in  plane  N. 
Case  I.    AO  is  drawn  from  point  A,  a  point  not  in  CD. 
Analysis  and  construction  (FiG.  31) : 
I.  To  prove  that  AO  lies  wholly  in  plane  N,  prove 
that  AO  coincides  with  a  line  that  does  lie  in 
plane  N. 
.'.  construct  AP  from  A  _L  CD  and  prove  that  AO 

coincides  with  AP. 

To  prove  that  AO  coincides  with  AP,  show  that 
AO  and  AP  are  both  J_  plane  M  from  A. 

Case  II.     AO  is  drawn  from  point  0  in  line  CD  (Fig.  32) . 
Analysis  and  proof  are  left  to  the  pupil. 

Ex.  1.  The  plane  perpendicular  to  the  edge  of  a  dihedral  angle 
from  any  point  P  contains  the  perpendiculars  from  P  to  the 
faces  of  the  dihedral  angle. 

Ex.  2.  Prove  that  the  perpendiculars  from  P  referred  to  in 
Ex.  1  form  an  angle  which  is  the  supplement  of  a  plane  of  the 
dihedral  angle. 


II. 


III. 


LINES  AND  PLANES 
TEST   FOR   PERPENDICULARS   TO    PLANES 


27 


38.  THEOREM  28.  If  two  intersecting  planes  are  per- 
pendicular to  a  third  plane,  their  intersection's  perpendicular 
to  that  plane. 


FIG.  33 

Hypothesis:  Planes  P  and  Q  _L  plane  M.  Planes  P  and 
Q  intersect  in  line  AB. 

Conclusion:    AB  _L  plane  M. 
Analysis  and  construction: 

I.  To  prove  AB  J_  plane  M,  prove  that  A B  coincides 

with  a  line  that  J_  plane  M. 
II.    /.  construct  AC  from  any  point  in  AB  _L  plane  M 

and  prove  that  AC  coincides  with  AB. 
III.  To  prove  that  AC  coincides  with  A  B,  prove  that  AC 
lies  in  both  plane  P  and  plane  Q. 

The  proof  is  left  to  the  pupil. 

Ex.  1.  Prove  Th.  28  by  drawing  the  construction  line  referred 
to  in  step  I  of  the  analysis  from  B  perpendicular  to  plane  M. 

Ex.  2.  If  three  non-parallel  planes  are  each  perpendicular  to 
a  third  plane,  their  intersections  are  parallel. 

Ex.  3.-  If  two  intersecting  planes  are  perpendicular  respectively 
to  two  intersecting  lines,  the  line  determined  by  the  planes  is 
perpendicular  to  the  plane  determined  by  the  lines. 

Ex.  4.  Two  dihedral  angles  with  their  edges  parallel  and  their 
faces  perpendicular  to  each  other  are  either  equal  or  supplemen- 
tary. 

Suggestion.     Pass  a  plane  J.  the  two  parallel  edges. 


28 


SOLID    GEOMETRY 
DETERMINATION    OF   PLANES 


39.  THEOREM  29.  Through  a  straight  line  oblique  to  a 
plane  one  plane  can  be  passed  perpendicular  to  the  plane 
and  only  one. 


FIG.  34 

Hypothesis:     Line  a  is  oblique  to  plane  M . 
Conclusion:     (1)  One  plane  can  be  passed  through  a  J_  M. 
(2)  Only  one  plane  can  be  passed  through  a  \_  M. 
Analysis  and  construction  (1) : 
I.  A  plane  J_  M  must  contain  a  line  _L  M. 
II.    .".  from  any  point  in  a  draw  PQ  J_  M  and  prove  the 
plane  of  a  and  PQ  J_  M. 

Analysis  (2) :  To  prove  plane  N  the  only  plane  through 
a  _L  M,  prove  that  there  is 

a.  Only  one  _L  from  P  to  M. 

b.  Only,  one  plane  containing  a  and  PQ. 

c.  Perpendiculars  to  M  from  all  points  in  a  lie  wholly 
in  AT. 

Let  the  pupil  give  the  proof. 

Ex.~~l.  In  Fig.  34  how  many  planes  could  be  drawn  perpen- 
dicular to  plane  M  and  containing  line  a  if  line  a  were  perpendicular 
to  plane  M? 

Ex.  2.  How  many  planes  can  be  drawn  to  contain  a  given 
point  and  perpendicular  to  a  given  plane?  How  would  these 
planes  be  obtained? 

Ex.  3.  Can  a  plane  be  drawn  through  a  given  point  perpen- 
dicular to  each  of  two  given  planes,  (1)  when  the  planes  intersect; 
(2)  when  the  planes  do  not  interesect?  Show  how  this  plane  is 
obtained  in  each  case... 


LINES   AND   PLANES  29 

PROPORTIONAL  SEGMENTS 

40.  THEOREM  30.  If  two  straight  lines  are  cut  by  three 
parallel  planes,  the  corresponding  segments  are  propor- 
tional. 


I       / 

FIG.  35 

Hypothesis:  The  parallel  planes  M,  N,  and  P  cut  the 
lines  h  and  k  at  points  A,  E,  C,  and  D,  G,  B  respectively. 

AE    DG 

Conclusion :     -=^  =  -7^5- 

£,C        LrD 

Analysis  and  construction: 

AE    DG 
I.   Po  prove  -£r  =  -(^ft>  prove  each  ratio  equal  to  a  third 

ratio. 

AE    AF       ,  DG    AF 
II.    /.  join  AB  and  prove  ——  and  ^=  — 

LOCI   IN   SPACE 

41.  As  in  plane  geometry,  so  in  solid  geometry  the  locus 
of  points  that  obey  one  or  more  requirements  consists  of  all 
points  that  satisfy  the  requirements  and  of  no  other  points. 
For  a  complete  proof  of  a  locus  theorem  it  is  necessary  to 
prove  that 

(1)  All  points  on  the  locus  obey  the  requirements. 

(2)  All  points  that  obey  the  requirements  are  on  the 
locus. 


30 


SOLID  GEOMETRY 
IMPORTANT  SPECIAL  CASES 


42.  THEOREM  31.  The  locus  of  points  equally  distant 
from  the  faces  of  a  dihedral  angle  is  the  half-plane  bisecting 
the  dihedral  angle. 


D 


Hypothesis:    A-BC-D  is  a  dihedral  angle  bisected  by  the 
plane  E. 

Conclusion:     Plane  E  is  the  locus  of  points  equally  dis- 
tant from  the  faces  M  and  N;  that  is, 

a.  Every  point  in  E  is  equally  distant  from  M  and  N. 

b.  Every  point  equally  distant  from  M  and  N  lies  in  E. 
Analysis  and  construction  for  a: 

I.  Let  P  be  any  point  in  plane  E. 
II.   /.  draw  PXJL-MandPY±N  and  prove  PX  =  PY 

III.  To  prove  PX  =  PY,  let  the  plane  of  PX  and  PY 

intersect  M  in  XZ,  and  N  in  FZ,  and  E  in  PZ 
and  prove  APXZmAPYZ. 

IV.  To  prove  APXZ&APYZ,  prove  ZPZX  =  ZPZ7. 
V.    .'.    prove  Z  PZX  and  Z  PZF  plane  angles  of  the 

dihedral  angles  A-BC-E  and  D-BC-E. 
VI.   /.  prove  plane  PXY  JL  5C. 

Analysis  and  construction  for  b: 

I.  Let  P  be  any  point  equally  distant  from  M  and  N. 
/.  let  PX  = 


LINES  AND   PLANES 


31 


II.  To  prove  that  P  lies  in  plane  E,  draw  plane  PBC 
and  prove  that  plane  PBC  bisects  the  dihedral 
angle  A-BC-D. 

III.    /.  let  the  plane  of  PX  and  PY  intersect  M  in  XZ, 
N  in  ZY,  and  PBC  in  PZ  and  prove  that 

(1)  Z  PZX  =  Z  PZF.    (Prove  APZX  ^  &PZY.) . 

(2)  A  PZX  and  PZY  are  plane  angles  of  the  dihedral 
angles. 

43.  THEOREM  32.  The  locus  of  points  equally  distant 
from  two  given  points  is  a  plane  bisecting  at  right  angles 
the  segment  joining  the  given  points. 


FIG.  37 

Hypothesis:     M  is  a  plane  bisecting  at  right  angles  the 
segment  joining  A  and  B. 

Conclusion:    M  is  the  locus  of  points  equally  distant 
from  A  and  B;  that  is, 

a.  Every  point  in  M  is  equally  distant  from  A  and  B. 

b.  Every  point  equally  distant  from  A  and  B  lies  in  M. 
Analysis  for  a: 

I.  Let  P  be  any  point  in  M. 

II.  To  prove  P  equally  distant  from  A  and  B,  join  PO 
and  prove  PO  J_  AB  (Plane  Geometry,  Th.  86). 

Analysis  for  b: 
I.  Let  Q  be  any  point  equally  distant  from  A  and  B; 

that  is,  let  QA=QB. 
II.  To  prove  Q  lies  in  M,  prove  QO  ±  AB  (see  Th.  12). 


32 


SOLID   GEOMETRY 


GENERAL   DISCUSSION 

44.  In  solid  geometry  two  kinds  of  loci  of  points  are  to 
be  considered: 

I.  Loci  of  points  fulfilling  one  requirement.     In  this  case 
the  locus  is,  in  general,  a  surface  or  a  number  of  surfaces. 

II.  Loci  of  points  fulfilling  two  requirements.     In  this 
case  the  locus  is,  in  general,  a  line  or  a  number  of  lines. 

If  three  requirements  are  given,  one  or  more  points  are, 
in  general,  determined. 

LOCUS   OF  POINTS   FULFILLING    ONE   REQUIREMENT 

45.  The   definitions   and   exercises   in   this   article   give 
examples  of  loci  of  points  fulfilling 

one  condition. 

The  surface  formed  when  a 
circle  is  revolved  about  a  diameter 
is  called  a  spherical  surface.  A 
spherical  surface  may  be  defined  as 
the  locus  of  points  in  space  at  a 
given  distance  from  a  given  point 
(Fig.  38). 

Ex.  1.  From  what  plane  locus  is  a 
spherical  surface  obtained?  How? 

The  surface  formed  when  one  of 
two  parallel  lines  revolves  about  the 
other  is  called  a  cylindrical  surface 
of  revolution.  It  is  the  locus  of 
points  in  space  at  a  given  distance 
from  a  given  line  (Fig.  39). 

Ex.  2.  From  what  plane  locus  may  a   cylindrical   surface  be 
obtained.     How? 

Ex.  3.  Find  the  locus  of  points  in  space  that   are  equally 
distant  from  two  parallel  planes. 

Ex.  4.  Find  the  locus  of  points  in  space  that  are  at  a  given 
distance  from  a  given  plane. 


FIG.  38 


FIG.  39 


LINES  AND   PLANES  33 

LOCI  OF  POINTS  FULFILLING  TWO  REQUIREMENTS 

46.  In  the  '  following  exercises  the  loci  required  are 
obtained  by  the  intersection  of  two  surfaces.  The  method 
of  treatment  is  indicated  in  the  suggestions  that  follow  Ex.  1. 

Ex.  1.  Find  the  locus  of  points  in  space  that  are  equally  dis- 
tant from  two  given  parallel  planes  and  also  equally  distant  from 
two  given  points. 

Solution: 

I.  All  points  equally  distant  from  two  given  parallel  planes  lie  in 
what  plane?     Call  this  plane  locus  I. 

II.  -All  points  equally  distant  from  two  given  points  lie  in  what 
plane?     Call  this  plane  locus  II. 

III.  Let  locus  I  intersect  locus  II  in  line  a.      .:   all  points  'that 
fulfill  both  requirements  lie  in  line  a. 

Discussion: 

I.  Ordinarily  the  required  locus  consists  of  one  line,  the  intersec- 
tion of  loci  I  and  II. 

II.  If  loci  I  and  II  coincide,  the  required  locus  is  a  plane. 

III.  If  loci  I  and  II  are  parallel,  the  problem  has  no  solution. 

Find  the  locus  of  points  in  a  given  plane  which  are  also 
Ex.  2.  Equally  distant  from  two  given  parallel  planes. 
Ex.  3.  Equally  distant  from  two  given  points  not  in  the  given 
plane. 

Ex.  4.  At  a  given  distance  from  a  second  given  plane. 

Ex.  5.  Equally  distant  from  two  other  given  intersecting  planes. 

Find  the  locus  of  points  in  space  which  fullfiill  the  require- 
ments given  in  Ex.  6-9. 

Ex.  6.  At  a  given  distance  from  a  given  plane  and  also  equally 
distant  from  two  given  points  not  in  the  given  plane. 

Ex.  7.  Equally  distant  from  two  given  points  A  and  B  and 
also  equally  distant  from  two  other  given  points  C  and  D. 

Ex.  8.  Equally  distant  from  three  given  points.  • 

Ex.  9.  Equally' distant  from  three  planes  that  meet  in  a  point. 

Ex.  10.  Equally  distant  from  two  given  points  A  and  B  and 
also  equally  distant  from  two  given  planes,  (1)  when  the  planes 
are  parallel,  (2)  when  the  planes  intersect. 


34  SOLID   GEOMETRY 

EXERCISES   INVOLVING   MISCELLANEOUS   LOCI 

47.  1.  Find  the  locus  of  points  in  space  equally  distant  from 
the  vertices  of  a  triangle.  Prove. 

2.  Find  the  locus  of  points  in  space  equally  distant  from  every 
point  on  a  circle.     Prove. 

3.  Determine  a  point  in  a  given  plane  which  is  equally  distant 
from  the  vertices  of  a  triangle.     Discuss  all  possibilities. 

4.  Determine  a  point  in  a  given  plane  that  is  equally  distant 
from  every  point  in  a  circle.     Discuss  all  possibilities. 

5.  Why  are  there  two  parts  to  Fig.  40? 

We  have  in  solid  geometry  not  only  loci  of  points,  but 
also  loci  of  lines  and  even  of  surfaces.  The  following  defini- 
tions and  exercises  illustrate  some  of  the  possibilities. 

The  surface  formed  when  one  of  two  intersecting  lines 
that  form  an  acute  angle  with  each  other  revolves  about  the 
other  as  an  axis  is  called  a  conical  surface  of  revolution. 
It  is  the  locus  of  lines  that  make  a  given  acute  angle 
with  a  given  line  at  a  given  point  in  the  line  (Fig.  40). 

The  line  used  as  an  axis  about  which  the  other  line  revolves 
is  not  shown  in  Fig.  40.  What  is  the  position  of  this  axis  ? 
What  does  the  figure  show  ? 

6.  What  is  the  locus  generated  when  the  lines 
are  at  right  angles? 

7.  What  is  the  locus  of  lines  that  are  parallel 
to  a  given  line  and  at  a  given  distance  from  it? 

8.  What  is  the  locus    of   lines  that    make  a 
given  angle  with  a  given  plane  at  a  given  point 
in  the  plane?    Discuss  various  possibilities. 

9.  What  is  the  locus  of  lines  that  pass  through  a 
fixed  point  and  are  parallel  to  a  fixed  plane? 

10.  Cut  out  a  rectangle  and  imagine  it  to  move  in  a  direction 
perpendicular  to  its  surface.    What  is  the  path  of  the  surface  of 
the  moving  rectangle? 

11.  Answer  the  preceding  question  using  a  circle  instead  of  a 
rectangle. 


LINES  AND   PLANES  35 

12.  Can  you  move  the  rectangle  referred  to  in  Ex.  10  so  that 
the  path  of  its  surface  will  be  different  from  that  obtained  in 
Ex.  10? 

13.  Answer  questions  10  and  12  for  various  kinds  of  triangles 
instead  of  for  a  rectangle. 

14.  Can  you  make  any  general  statement  concerning  the  path 
of  (1)  a  moving  point;   (2)  a  moving  line;    (3)  a  moving  surface? 
Suppose  in  each  case  that  the  moving  point,  line,  or  surface  obeys 
one  or  more  requirements. 

15.  What  is  the  path  of  a  moving  solid? 

PROJECTION 

48.  Objects  are  represented  or  pictured  on  surfaces  by 
means  of  projection.  Projections  are  of  two  kinds,  parallel 
and  central. 

Fig.  41  illustrates  parallel  projection.  Parallel  rays 
through  the  points  A,  B,  C,  D  intersect  the  plane  M  in  the 
points  A',  B',  C't  D'.  A'B'C'D'  is  the 
projection  of  A  BCD  on  the  plane  M. 
The  projecting  lines  may  make  any  angle 
with  the  plane.  The  plane  M  may  be 
replaced  by  any  surface  whatsoever. 
A  shadow  made  by  the  sun  is  an  ex-  A  ~B 
ample  of  parallel  projection.  FIG.  41 

By  the  projection  of  a  point  on  a  plane  is  meant  the  inter- 
section of  the  projecting  ray  and  the  plane. 

By  the  projection  of  a  line  on  a  plane  is  meant  the  locus 
of  the  projections  of  all  of  its  points. 

If  the  projecting  rays  are  perpendicular  to  the  plane,  the 
projection  is  said  to  be  orthogonal. 

If  the  rays  start  from  a  common  origin,  instead  of  being 
parallel,  the  projection  is  said  to  be  central.  A  shadow  made 
by  a  candle  is  an  example  of  central  projection. 

Hereafter  when  the  word  "projection"  is  used  orthogonal 
projection  is  intended. 


30  SOLID   GEOMETRY 

49.  THEOREM  33.     The  projection  of  a  straight  line  not 
perpendicular  to  a  plane  upon  that  plane  is  a  straight  line. 


FIG.  42 

Analysis  and  construction: 
I.  Prove  that  the  projection  of  a  on  M  coincides  with 

a  line  that  is  straight. 

II.    .'.   construct  plane  N  through  a  J_  M,  intersecting 
M  in  b.     Prove  that  b  is  the  projection  of  a  on  M. 

III.  .'.  prove  that  b  contains  the  projections  of  all  points 

in  a. 

IV.  /.  prove  that  _Ls  to  M  from  all  points  in  a  lie  in  N. 

COR.  If  a  straight  line  is  oblique  to  a  given  plane,  its 
projection  upon  that  plane  is  the  intersection  of  the  given 
plane  with  a  plane  through  the  line  perpendicular  to  the 
plane. 

The  projection  of  a  segment  on  a  plane  is  the  segment 
between  the  feet  of  the  perpendiculars  drawn  to  the  plane 
from  the  extremities  of  the  given  segment. 

Ex.  1.  If  a  segment  is  parallel  to  a  plane,  it  is  parallel  and 
equal  to  its  projection  on  the  plane. 

Ex.  2.  Can  the  projection  of  a  curve  upon  a  plane  be  a  straight 
line?  How? 

Ex.  3.  Can  the  projection  of  a  curve  upon  each  of  two  inter- 
secting planes  be  straight  lines?  How? 

Ex.  4.  If  two  lines  are  parallel,  their  projections  on  the  same 
plane  coincide  or  are  parallel. 

Ex.  5.  The  projections  of  a  segment  upon  two  parallel  planes 
are  parallel  and  equal. 


LINES  AND   PLANES 


37 


50.  THEOREM  34.  The  acute  angle  that  a  straight  line 
makes  with  its  own  projection  on  a  plane  is  the  least  angle 
that  it  makes  with  any  line  of  the  plane. 


FIG.  43 

Analysis  and  construction:  To  prove  £BAC<Z.BAD, 
construct  BC  J_  M  from  B,  any  point  in  /,  and  intersecting 
line  x  at  C.  Make  AD  on  line  y  equal  to  AC.  Join  BD. 
Prove  BC<BD  (see  Plane  Geometry,  Th.  60). 

An  angle  that  a  straight  line  makes  with  its  own  projec- 
tion on  a  plane  is  called  the  inclination  of  the  line  to  the 
plane  or  the  angle  of  the  line  and  the  plane. 

Ex.  1.  If  a  line  is  oblique  to  a  plane,  what  is  the  largest  angle 
that  it  makes  with  any  line  of  the  plane?  Why? 

Ex.  2.  If  a  straight  line  intersects  two  parallel  planes,  it  makes 
equal  angles  with  them. 

51.  THEOREM  35.  The  length  of  the  projection  of  a 
given  segment  on  a  plane  is  the  length  of  the  segment 
multiplied  by  the  cosine  of  the  angle  of  inclination  of  the 
line  to  the  plane  (Fig.  44). 

Suggestion.  Show  that  A  B  -  A  C  cos  /.BAG.  :.  XY  =  ACcos  /.BAG. 

Exercise.    Using  the  tables,  find  the  lengths  of  the  projections  of 
the  following  segments.    The 
lengths  and  the  angles  made 
with  the  planes  are  given: 

a.  15  in.,   42°. 

b.  27  in.,   38°. 

c.  36  in.,   67°.  ^ ^---g M' 

d.  23  in.,   52°.  FIG.  44 


38  SOLID   GEOMETRY 

* 

POLYHEDRAL  ANGLES 

62.  If  a  ray  has  a  fixed  origin  and  moves  so  as  continually 
to  intersect  the  perimeter  of  a  fixed  polygon  not  in  the  plane 
of  its  origin,  the  ray  is  said  to  generate  a  polyhedral  angle. 

In  Fig.  45,  0-ABCD  etc.  is  a  poly- 
hedral angle.  0  is  the  fixed  origin  of 
the  ray  and  is  called  the  vertex  of  the 
polyhedral  angle.  ABCD  etc.  is  the  fixed 
polygon.  OA,  OB,  etc.,  are  called  the 
edges  of  the  polyhedral  angle.  The  planes 
AOB,  BOC,  etc.,  are  called  its  faces. 
The  angles  AOB,  BOC,  etc.,  made  by 
two  consecutive  edges  are  called  the  face 
angles  of  the  polyhedral  angle.  The  dihedral  angles  made 
by  two  consecutive  faces  are  called  its  dihedral  angles. 

The  parts  of  a  polyhedral  angle  are  its  face  angles,  its 
faces,  and  its  dihedral  angles. 

A  polyhedral  angle  is  said  to  be  convex  or  concave  accord- 
ing as  the  fixed  polygon  is  convex  or  concave.  Only  convex 
polyhedral  angles  will  be  considered. 

A  polyhedral  angle  with  three  faces  is  called  a  trihedral 
angle. 

The  size  of  a  polyhedral  angle  depends  upon  the  spread 
of  the  planes  forming  the  faces  of  the  polyhedral  angle.  The 
measure  of  the  polyhedral  angle  is  too  difficult  for  this  book. 

A  polyhedral  angle  is  one  kind  of  a  solid  angle.  Another 
kind  of  a  solid  angle  is  the  conical  angle.  A  conical  angle  is 
generated  by  a  ray  that  has  a  fixed  origin  and  moves  so  as 
continually  to  intersect  a  fixed  closed  curve. 

If  the  face  angles  and  the  dihedral  angles  of  one  poly- 
hedral angle  are  equal  respectively  to  the  corresponding  parts 
of  the  other,  and  arranged  in  the  same  order,  the  polyhedral 
angles  are  said  to  be  congruent. 


LINES  AND   PLANES  39 

If  the  face  angles  and  the  dihedral  angles  of  one  poly- 
hedral angle  are  equal  respectively  to  the  corresponding 
parts  of  the  other,  but  arranged  in  the  opposite  order,  the 
polyhedral  angles  are  said  to  be  symmetric. 

Ex.  1.  Construct  out  of  cardboard  two  congruent  trihedral 
angles;  also  two  symmetric  trihedral  angles. 

Ex.  2.  If  the  edges  of  one  polyhedral  angle  are  extended  through 
the  vertex,  a  polyhedral  angle  is  formed  which  is  symmetric  to  the 
given  polyhedral  angle. 

63.  THEOREM  36.  Any  face  angle  of  a  trihedral  angle  is 
less  than  the  sum  of  the  other  two. 


Hypothesis:    A-BCD  is  a  trihedral  angle  with  the  face 
angles  1,  2,  3,  and  the  edges  AY,  AZ,  and  AW. 

Conclusion:     ZKZ2-hZ3. 

Case  A.     When  Z  1  =  either  Z 2  or  Z 3. 

Case  B.     When  Z  1  <  either  Z  2  or  Z  3. 

CaseC.     When  Zl>either  Z2  or  Z3. 

Proof  is  required  for  Case  C  only.     Why? 

Analysis  and  construction: 

I.  To  prove  Z  1<  Z2+ Z 3,  construct  Z.XAW  in  Zl 
so  that  ZX,W=Z3  and  prove  ZYAX<Z2. 
II.  .*.  lay  off  equal  segments  AE  and  AC  on  AX  and 
AZ  respectively.  Pass  any  plane  through  points 
TTand  C  cutting  A W  and  AY  at  D  and  B  respec- 
tively. Prove  EB<BC  (Plane  Geometry,  Th.  60). 

III.   /.  since  BD<BC+CD,  prove  DE  =  CD. 


f  \JfiS    ^V 

40  SOLID  GEOMETRY 

54.  THEOREM  37.     The  sum  of  the  face  angles  of  any 
polyhedral  angle  is  less  than  four  right  angles. 


FIG.  47 

Hypothesis:     A-BCDE  etc.  is  any  polyhedral  angle  with 
face  angles  1,  2,  3,  etc. 

Conclusion:     Z1  +  Z2  +  Z3  +  etc.   <  4  rt.   A . 
Analysis  and  construction: 
I.  To  prove  Z1+Z2  +  Z3  +  etc.  <  4  rt.  A ,  compare 

them  with  angles  that  are  equal  to  4  rt.  A  . 
II.    .'.  construct  plane  BCDE  etc.  cutting  all  the  edges 
of  A-BCDE  etc.,  but   not   containing  point  A. 
Take  0  any  point  in  polygon  ABCD   etc.,  join 
BO,  CO,  DO,  etc.,  and  prove  the  sum  of  the  A 
about  A<  the  sum  of  the  A  about  0. 
III.    /.  compare  the  sum  of  the  A  of  the  A  A  with  the 
sum  of  the   A   of  the  OA  and  the  sum  of  the 
base  A  of  A  A  with  the  sum  of  the  base  A  of  OAN 
Outline  oj  proof: 

I.  Sum  of  the  A  of  A  A  =sum  of  the  A  of  OA. 
II.  ABCA+Z.ACD>ABCD. 

Z  CD  A  +  Z  A  DE  >  Z  CDE,  etc. 
/.  sum  of  the  base  A  of  A  A  >sum  of  base  A  of  0  A. 
III.    .*.  sum  of  vertex  A  of  A  A  <-sum  of  vertex  A  of  0  A . 

(See  Plane  Geometry,  As.  34.) 
.'.  Z1+Z2  +  Z3  +  etc.<4  rt.  A  . 

Exercise.    If  a  trihedral  angle  has  two  face  angles  right  angles, 
two  of  its  dihedral  angles  are  right  dihedral  angles. 


LINES  AND   PLANES 


41 


65.  THEOREM  38.  If  two  trihedral  angles  have  the  three 
face  angles  of  one  equal  to  the  three  face  angles  of  the  other, 
the  dihedral  angles  opposite  equal  face  angles  are  equal. 


II. 


Fig.  48 


Analysis  and  construction: 

I.  To  prove  dihedral  angle  OA  equals  dihedral  angle 
O'A',  prove  their  plane  angles  equal. 

take  O'A'  =  0'B'  =  0'C'  =  OA=OB  =  OC.  Join 
AB,  BC,  CA  and  A'B't  B'C,  C'A'.  Take  OX 
in  OA  equal  to  O'X'  in  O'A'.  Draw  XV  J_  OA 
at  X  in  AOB,  XZ  J_  OA  at  X  in  AOC.  Let  XY 
and  XZ  meet  AB  and  AC  at  Y  and  Z  respectively. 
X'Y'  and  AT'Z'  are  similarly  drawn  in  the  faces 
of  angle  0'.  Join  YZ  and  Y'Z'.  .  Prove  Z  YXZ 
=  Z  Y'X'Z'. 

prove  AXYZ^AX'Y'Z'. 
prove  XY=X'Y',  XZ  =  X'Z't  YZ=Y'Z'. 
V.  To  prove  XY  =  X'Y';  prove  AAXY  giAA'X'Y'. 
VI.  To  prove  AAXY  a  AA'AT'Y',  prove     .... 
Let  the  pupil  complete  the  analysis  and  give  proof. 
In  the    same  way  the   other  dihedral   angles   may  be 
proved  equal. 

Question.  It  is  necessary  for  the  proof  that  XY  does  not  meet  AB 
on  BA  extended.  How  does  the  construction  provide  for  this? 

COR.  If  the  three  face  angles  of  one  trihedral  angle  are 
equal  to  the  three  face  angles  of  another,  the  trihedral 
angles  are  congruent  if  the  parts  are  arranged  in  the  same 
order  and  symmetric  if  arranged  hi  the  opposite  order. 


III. 
IV. 


42  SOLID   GEOMETRY 

SUMMARY  AND   SUPPLEMENTARY  EXERCISES 
56.   SUMMARY  OF  IMPORTANT  POINTS  IN  CHAPTER  I 

A.  DETERMINATION  OF  LINES  AND  PLANES: 

I.  Only  one  plane  can  be  drawn  containing 

a.  Three  given  points  (§3). 

b.  One  given  line  and  a  given  point  not  in  the 

line  (§3). 

c.  Two  given  intersecting  lines  (§3). 

d.  Two  given  parallel  lines  (§3). 

e.  One  given  point  and  be  perpendicular  to  a 

given  line  (§18). 

/.  One  given  line  oblique  to  a  given  plane  and 
be  perpendicular  to  the  given  plane  (§39). 

II.  Only  one  line  can  be  drawn  containing  a  given 
point  and 

a.  Be  perpendicular  to  a  given  line  if  the  point 

is  not  on  the  line  (§15). 

b.  Be  perpendicular  to  a  given  plane  (§20). 

c.  Be  parallel  to  a  given  line. 

d.  Contain  a  second  given  point. 

III.  A  given  line  lies  in  a  plane  if  it  contains 

a.  A  point  in  the  plane  and  is  parallel  to  a  line 

that  is  parallel  to  the  plane  (§10)'. 

b.  A  point  in  the  plane  and  is  perpendicular  to 

a  line  that  is  perpendicular  to  the  plane 
(§19). 

c.  A  point  in  one  of  two  perpendicular  planes 

and  is  perpendicular  to  the  other  (§37). 

d.  Two  points  in  the  plane  (§3). 

B.  TESTS: 

I.  To  prove  planes  parallel  to  planes,  prove  that 

a.  They  cannot  meet  (§4). 

b.  Two  intersecting  lines  in  one,  etc.  (§12). 

c.  They  are  perpendicular  to  the  same  line  (§27) . 


LINES  AND   PLANES  43 

IF    To  prove  a  line  and  a  plane  parallel,  prove  that 

a.  They  cannot  meet  (§6). 

b.  The  plane  contains  a  line  that  is  parallel  to 

the  given  line  (§9). 

ft  I.  To  prove  two  lines  parallel,  prove  that 

a.  They  are  in  the  same  plane  and  do  not  meet 

(§7). 

b.  One  is  parallel  to  a  given  plane  and  the  other 

is  the  intersection  of  the  given  plane  with 
a  plank  containing  the  given  line  (§10). 

c.  They  are  the  intersections  of  two  parallel 

planes  cut  by  a  third  (§11). 

d.  They  are  parallel  to  the  same  line  (§§13,  25). 

e.  They  are  perpendicular  to  the  same  plane 

(§25). 

IV.  To  prove  a  line  perpendicular  to  a  given  plane, 

prove  that 

a.  It  is  perpendicular  to  two  lines  of  the  plane 

at  their  intersection  (§16). 

b.  It  is  parallel  to  a  line  that  is  perpendicular  to 

the  plane  (§24). 

,  c.  It  is  perpendicular  to  one  of  two  parallel 
planes  (§26). 

d.  It  lies  in  one  of  two  perpendicular  planes  and 

is  perpendicular  to  their  intersection  (§36). 

e.  It  is  the  intersection  of  two  planes  that  are 

each  perpendicular  to  the  given  plane  (§38) . 

V.  To  prove  two  lines  perpendicular,  prove  that  one 
is  perpendicular  to  a  plane  containing  the  other, 
at  a  point  in  the  other  (§16). 
VI.  To  prove  two  planes  perpendicular,  prove  that 

a.  One  contains  a  line  perpendicular  to  the  other 

(§35). 

b.  The  dihedral  angle  formed  is  a  right  dihedral 

angle  (§34). 


44 


SOLID   GEOMETRY 


EXERCISES   IN   CONSTRUCTION 

57.  1.  Show  how  to  construct  a  plane  perpendicular  to  a  given 
line  at  a  given  point  in  the  line. 

2.  Show  how  to  construct  a  plane  perpendicular  to  a  given 
line  and  containing  a  given  point  without  the  line. 

3.  From  a  point  in  a  given  plane 
construct    a    line    perpendicular    to 
the  plane. 

A  nalysis  and  construction  (Fig.  49) : 
I.  Draw  OB  any  line  in  M  through 

0.    Construct  plane  N  J_  OB 

at  0.     Draw  A  0  in  TV  J_  the 

intersection  of  M  and  N  at  0. 

'  II.  Prove  N  J_ 


FIG.  49 
Use  Ths.  25  and  26. 

4.  From  a  point  without  a  given  plane  construct  a  line  per- 
pendicular to  the  given  plane. 

Analysis  and  construction   (Fig.  50): 
I.  AO  is  to  be  in  a  plane  that  is 

J_   M  and   _L  the  intersection 

of  this  plane  and  M. 
II.  Draw  AB  from  A  J_  x  any  line  in 

M.    Draw  line  c  from  £,  in  M, 

_L  #.       Draw    AO    J_    line   c 

from  A. 

5.  Show  that  in  Ex.  3  and  4  (M 
may  be  proved  _L  Af  by  proving  it  JL 
to  two  lines  in  plane  M . 

6.  Show  that  Ex.  4  may  be  solved  as  follows:    Draw  three  equal 
segments  from  A  to  the  plane,  meeting  the  plane  at  B,  C,  and  D. 
Find  0,  the  center  of  the  circle  through  B,  C,  and  D.     Join  AO. 

7.  Through  a  given  line  pass  a  plane  perpendicular  to  a  given 
plane.    How  many  such  planes  are  possible?   Discuss  various  cases. 

8.  Through  a  given  point  construct  a  plane  that  shall  be  per- 
pendicular to  each  of  two  given  planes:    (1)  when  the  planes 
intersect;    (2)  when  the  planes  are  parallel. 

9.  Construct  a  plane  through  a  given  point  and  parallel  to  a 
given  plane.     Is  more  that  one  such  plane  possible? 


E 
FIG.  50 


LINES  AND   PLANES 


45 


FIG.  51 


10.  Through  a  given  point  in  one  of  two  planes  draw  a  line 
parallel  to  the  other  plane:  (1)  when  the  planes  intersect;  (2)  when 
the  planes  are  parallel. 

11.  Through  a  given  point   in  space  draw  a  line  parallel  to 
each  of  two  given  planes   (1)  when  the  planes  intersect;   (2)  when 
the  planes  are  parallel. 

12.  Construct  a  common  perpen- 
dicular to  two  skew  lines  (Fig.  51). 

Analysis  and  construction: 
I.  If  d  is  the  required  perpendic- 
ular, d  must  (1)  be  _|_  b,  (2)  be 
JL  a  line  as  c  which  is  ||  a  and 
intersects  6,  (3)  lie  in  same 
plane  as  a  and  c. 

II.  Draw  plane  M  through  b  ||  a.     Line  a  will  be  ||  any  line  in 

M  that  lies  in  the  same  plane  as  a. 

III.    .'.  draw  through  a  plane  TV  _L  M ,  cutting  M  in  line  c.     Let  c 
cut  b  in  point  .  I . 

IV.    /.  drawd  J_  Mat  A. 

13.  Show  that  only  one  common  perpendicular  can  be  drawn 
to  two  skew  lines. 

Suggestion  (Fig.  51).  Show  that  if  any  other  line  (as/)  is  J_  lines 
a  and  b,  and  if  /  intersects  line  a  at  point  X,  we  could  have  two  perpen- 
diculars to  M  from  X. 

14.  Show  that  the  common  perpendicular  to  two  skew  lines  is 
the  shortest  distance  between  them. 

Suggestion.     Line  /  >  XY  and  .'.    >  d. 

15.  Through  a  given  point  draw  a  straight  line  that  will  inter- 
sect each  of  two  given  skew  lines. 

Suggestion.  Draw  two  planes  each  determined  by  the  given  point 
and  one  of  the  given  lines. 

16.  Given  three  non-parallel  non-intersecting  straight  lines. 
Show  that  any  number  of  straight  lines  can  be  drawn  that  will 
intersect  the  three. 

Suggestion.     Construct  a  pencil  of  planes  through  one  of  the  lines. 

17.  Construct  a  pair  of  parallel  planes  so  that  each  shall  con- 
tain one  of  two  skew  lines. 


46  SOLID   GEOMETRY 

EXERCISES  INVOLVING  LOCI  AND  CONCURRENT  LINES 

58.  1.  Prove  that  the  planes  bisecting  the  dihedral  angles  of  a 
trihedral  angle  are  concurrent. 
Analysis: 

I.  Let  a,  b,  and  c  be  the  faces  of  dihedral  angle  0. 
II.  To  prove  the  bisectors  of  the  dihedral  angles  concurrent, 
prove  that  the  bisector  of  the  angle  made  by  a  and  b  and 
that  made  by  b  and  c  meet  in  a  line  and  that  this  line  lies 
in  the  bisector  of  the  angle  made  by  a  and  c. 
III.    .'.  prove  that  every  point  in  this  line  is  equally  distant  from  a 
and  c. 

2.  What  locus  is  obtained  by  the  intersection  of  three  planes 
referred  to  in  Ex.  1? 

3.  Find  the  locus  of  points  in  space  that  are  equally  distant 
from  three  given  planes  that  intersect  each  other  in  three  parallel 
lines. 

4.  Prove  that  the  locus  of  points  in  space  that  are  equally 
distant  from  two  intersecting  lines  is  the  plane  that  is  perpen- 
dicular to  the  plane  of  the  lines  and  that  contains  the  bisectors  of 
the  angles  formed  by  the  lines. 

5.  Prove  that  the  planes  perpendicular  to  the  faces  of  a  tri- 
hedral angle  and  passing  through  the  bisectors  of  the  face  angles 
are  concurrent.    What  locus  is  obtained? 

6.  Prove  that  the  planes  that  pass  through 
the  edges  of  a  trihedral  angle  and  the  bisec- 
tors of  the  opposite  face  angles  are  concur- 
rent (Fig.  52). 

Analysis  and  construction: 
I.  To  prove  OA  Y,  OBZ,  and  OCX  concur- 
rent, prove  that  they  have  two  points 

•  I/ IG 

in  common. 
II.    .'.  make  OA  =  OB  =  OC  and  prove  that  OA  Y,  OBZ,  and  OCX 

intersect  plane  ABC  in  concurrent  lines. 
III.    /.  prove  A  Y,  BZ,  and  CX  the  medians  of  A  ABC. 

7.  Prove  that  the  planes  that  are  perpendicular  to  the  chords 
of  a  circle  at  their  midpoints  are  concurrent. 

8.  What  locus  is  obtained  by  the  intersection  of  the  three 
planes  referred  to  in  Ex.  7?    Give  proof. 


LINES  AND   PLANES  47 

EXERCISES   INVOLVING   PROJECTIONS 

69.  1.  If  two  parallel  lines  are  oblique  to  a  plane,  they  make 
equal  angles  with  the  plane. 

2.  Two  equal  segments  from  a  point  to  a  plane  have  equal 
projections  on  the  plane  and  make  equal  angles  with  the  plane. 

3.  If  a  segment  is  oblique  to  a  plane,  it  is  longer  than  its  pro- 
jection on  the  plane. 

4.  If  two  unequal  segments  are  drawn  from  a  point  to  a  plane, 
which  has  the  longer  projection  on  the  plane?     Give  proof.    Which 
makes  the  greater  angle  with  the  plane?     Give  proof. 

5.  If  a  straight  line  is  perpendicular  to  one  of  two  intersecting 
planes,  its  projection  on  the  other  is  perpendicular  to  the  inter- 
section of  the  two  planes. 

6.  If  a  right  angle  has  one  side  parallel  to  a  plane,  its  projec- 
tion on  the  plane  is  a  right  angle.     When  is  this  not  true? 

7.  When  is  the  projection  of  a  parallelogram  on  a  given  plane 
a  parallelogram?    Why? 

8.  Answer  the  question  in  the  preceding  exercise  for  a  rec- 
tangle instead  of  a  parallelogram. 

9.  The  sides  of  an  isosceles  triangle  make  equal  angles  with 
any  plane  containing  its  base. 

10.  Parallel  segments  are  proportional  to  their  projections  on 
the  same  plane. 

11.  If  the  projection  of  a  given  line  upon  each  of  two  inter- 
secting planes  is  a  straight  line,  the  given  line  is,  in  general,  a 
straight  line. 

12.  A  point  is  12  in.  from  a  given  plane.    What  is  the  length 
of  a  projection  of  a  segment  13  in.  long  drawn  from  the  point  to 
the  plane? 

13.  What  is  the  length  of  the  projection  of  a  segment  5  in.  long 
on  a  plane  if  the  segment  makes  an  angle  of  30°  with  the  plane? 
If  it  makes  an  angle  of  60°  with  the  plane?     If  it  makes  an  angle 
of  45°  with  the  plane?    Trigonometry  tables  are  not  necessary. 


CHAPTER   II 
PROPERTIES  OF  POLYHEDRONS,  CYLINDERS,  AND  CONES 

SOLIDS   IN   GENERAL 

60.  An  inclosed  portion  of  space  is  called  a  solid. 
Solids  are  bounded  or  inclosed  by  surfaces. 

If  any  solid  is  intersected  by  a  plane,  the  figure  formed 
by  the  intersections  of  the  plane  and  the  boundaries  of  the 
solid  is  called  a  section  of  the  solid. 

POLYHEDRONS  IN  GENERAL 

61.  In  general,  a  solid  all  of  whose  bounding  surfaces  are 
planes  is  called  a  polyhedron  (Fig.  53). 

The  lines  of  intersection  of 
the  bounding  planes  are  called 
the  edges  of  the  polyhedron. 

The  points  of  intersection  of 
the  edges  are  called  the  vertices 
of  the  polyhedron. 

The  polygons  bounded  by  the  edges  of  the  polyhedron 
are  called  the  faces  of  the  polyhedron. 

SOME   ELEMENTARY   SURFACES 

62.  We  have  seen  that  sometimes  a  surface  is  the  locus 
of  a  point  or  of  a  line  that   obeys  certain  requirements. 
The  surfaces  studied  in  solid  geometry  are  of  this  nature. 

We  shall  study  the  following  special  surfaces : 
If  a  straight  line  moves  so  as  continually  to  intersect  a 
given  straight  line,  and  at  the  same  time  to  remain  parallel 
to  a  third  straight  line  not  in  the  plane  of  the  second,  the 
surface  generated  is  a  plane  surface. 


POLYHEDRONS,  CYLINDERvS,  AND  CONES         49 


A  plane  surface,  or  merely  a  plane,  may 
be  defined  from  its  fundamental  characteris- 
tic (§3)  as  a  surface  such  that  if  any  two 
points  in  it  are  chosen  the  straight  line  pass- 
ing through  these  points  will  lie  wholly  in  the 
surface. 

If  a  straight  line  moves  so  as  continually 
to  intersect  a  chain  of  straight-line  segments 
that  lie  in  one  plane,  and  at  the  same  time 
to  remain  parallel  to  a  fixed  straight  line  not 
in  this  plane,  the  surface  generated  is  called 
a  prismatic  surface  (Fig.  54). 

If  a  straight  line  moves  so  as  continually 
to  intersect  a  fixed  plane  curve,  and  at  the 
same  time  to  remain  parallel  to  a  fixed  straight 
line  not  in  the  plane  of  the  curve,  the  surface 
generated  is  called  a  cylindrical  surface 
(Fig.  55). 


If  a  straight  line  moves  so  as  con- 
tinually to  intersect  a  chain  of  straight- 
line  segments  that  lie  in  one  plane,  and 
at  the  same  time  to  pass  through  a  given 
point  not  in  this  plane,  the  surface  gen- 
erated is  called  a  pyramidal  surface  (Fig. 
56).  The  fixed  point  is  called  the  vertex 
of  the  surface. 


If  a  straight  line  moves  so  as  continually 
to  intersect  a  fixed  plane  curve,  and  at  the 
same  time  to  pass  through  a  given  point  not 
in  the  plane  of  the  curve,  the  surface  gener- 
ated  is  called  a  conical  surface  (Fig.  57). 
The  fixed  point  is  called  the  vertex  of  the 
surface. 


FIG.  54 


FIG.  55 


FIG.  57 


50  SOLID  GEOMETRY 

The  prismatic  and  the  pyramidal  surfaces  are  examples 
of  plane  surfaces.  The  cylindrical  and  the  conical  surfaces 
are  special  kinds  of  curved  surfaces.  A  curved  surface  is  a 
surface  no  part  of  which  is  plane. 

NOTE.  There  are  many  kinds  of  curved  surfaces  besides  those 
mentioned  above.  If  a  circle  is  revolved  about  a  diameter,  a  spheri- 
cal surface  is  formed.  We  shall  study  spherical  surfaces  later.  The 
reflector  of  a  locomotive  headlight  is  a  surface  formed  by  the  revolution 
of  a  parabola. 

63.  In  the  surfaces  defined  above  the  moving  line  is  called 
the  generator.  The  series  of  straight-line  segments  or  the 
curve  is  called  the  director.  In  Figs.  54-57,  g  is 
the  generator  and  d  is  the  director. 

When  the  director  is  a  closed  polygon  or  a 
closed  curve  and  the  generator  moves  com- 
pletely around  the  director,  the  surface  is  said 
to  be  a  closed  surface  (Fig.  58) . 

If  the  director  is  a  convex  polygon  or  curve, 

the  surface  generated  is  said  to  be  a  convex 

.  FIG.  58 

surface. 

The  generator  in  any  position  is  called  an  element  of  the 
surface..  The  edges  of  a  prismatic  or  pyramidal  surface  are 
the  elements  that  pass  through  the  points  of  intersection  of 
the  segments  of  the  series  that  form  the  director. 

Pyramidal  and  conical  surfaces  each  consist  of  two  parts 
called  nappes.  The  generator  of  a  pyramidal  or  conical  sur- 
face is  a  line,  not  a  ray.  If,  in  Figs.  56,  57,  and  58,  0  is 
the  fixed  point  through  which  the  generator  passes,  the  part 
below  point  0  generates  one  part  or  nappe  of  the  surface 
formed,  while  the  part  above  point  0  generates  the  other 
part  or  nappe. 

The  two  nappes  of  a  closed  convex  pyramidal  surface 
contain  polyhedral  angles.  The  edges  of  one  are  prolonga- 
tions of  the  edges  of  the  other.  Are  the  polyhedral  angles 
equal  or  symmetric?  Why? 


POLYHEDRONS,  CYLINDERS,  AND  CONES         51 

If  a  plane  cuts  all  the  elements  of  a  closed  surface,  the 
figure  formed  by  the  intersection  of  the  plane  and  the  sur- 
face is  called  a  transverse  section  of  the  surface. 

A  transverse  section  formed  by  a  plane  perpendicular  to 
the  elements  of  a  prismatic  or  cylindrical  surface  is  called  a 
right  section  of  the  surface  (VWXYZ,  Fig.  60). 


PRISMS 
DEFINITIONS 

64.  An  unlimited  closed  prismatic  surface  is  said  to 
inclose  a  prismatic  space. 

A  solid  bounded  by  a  closed  pris- 
matic surface  and  two  parallel  trans- 
verse sections  is  called  a  prism 
(Figs.  59  and  60).  It  is  a  portion 
of  a  prismatic  space. 

The  two  transverse  sections  are 
called  the  bases  of  the  prism.  The  FIG-  59 

prismatic  surface  and  its 
edges  are  called  respec- 
tively the  lateral  surface 
and  lateral  edges  of  the 
prism. 

The  perpendicular 
distance  between  the 
bases  of  a  prism  is  called  — 
the  altitude  of  the  prism. 

Since  the  form  of  a  prismatic  space  depends  upon  the 
form  of  its  right  section,  prisms  may  be  named  from  the 
form  of  their  right  sections,  thus:  A  prism  will  be  called  a 
square  prism  if  its  right  section  is  a  square. 

Exercise.  Make  of  wood  or  cardboard  a  square  prism  in 
which  the  edges  are  not  perpendicular  to  the  base.  Is  the  base 
a  square? 


FIG.  60 


52  SOLID  GEOMETRY 

GENERAL   PROPERTIES    OF   PRISMS 

65.  THEOREM  39.     Two  parallel  transverse  sections  of 
a  prismatic  space  are  congruent. 


FIG.  61 

Analysis: 

I.  To   prove   s  ^sf  prove   the  sides  and  angles  equal 

in  the  same  order. 

II.  To  prove  AB  =  A'Bf,  prove  ABB' A'  a  O. 
III.  To  prove    /.ABC=  ^A'B'C',  prove  AB\\A'B'  and 
BC\\B'C'. 

66.  The  following  theorems  are  corollaries  of  Th.  39  and 
of  the  definition  of  a  prism. 

COR.  I.     The  bases  of  a  prism  are  congruent  polygons. 

COR.  II.  The  lateral  edges  of  a  prism  are  parallel  and 
equal. 

COR.  III.  The  lateral  faces  of  a  prism  are  parallelo- 
grams. 

COR.  IV.     All  right  sections  of  a  prism  are  congruent. 

Ex.  1.  Is  it  possible  for  only  one  lateral  face  of  a  prism  to  be 
a  rectangle?  Illustrate  by  a  model. 

Ex.  2.  Is  it  possible  for  only  two  lateral  faces  of  a  prism  to  be 
rectangles?  Prove. 

Ex.  3.  Every  pair  of  lateral  edges  of  a  prism  determines  a 
plane  parallel  to  each  of  the  other  lateral  edges. 

Ex.  4.  If  two  intersecting  planes  each  contain  one  and  only 
one  lateral  edge  of  a  prism,  their  intersection  is  parallel  to  the 
other  lateral  edges. 


POLYHEDRONS,  CYLINDERS,  AND  CONES          53 


FIG.  62 


67.  That  part  of  a  prism  included 
between   one   base    and  a   transverse 
section  oblique  to  the  base  is  called  a 
truncated  prism  (Fig.  62). 

SPECIAL  PRISMS* 

68.  A  prism  is  said  to  be  a  right  prism  if  its  base  is  a  right 
section  (Fig.  63).     A  prism  that  is  not  a  right  prism- is 
called  an  oblique  prism  (see  Fig.  59). 

THEOREM  40.     The  lateral  faces  of  a 
right  prism  are  rectangles. 

THEOREM  41.    The  lateral  faces  of  a 
right  prism  are    perpendicular    to    the 

base-  PIG.  63 

69.  A  prism  is  said  to  be  regular  if  it  is  a  right  prism  whose 
base  is  a  regular  polygon. 

70.  A  prism  is  called  a  parallelepiped  if  its  bases  are 
parallelograms  (Fig.  64). 

THEOREM  42.  Any  two  oppo- 
site faces  of  a  parallelepiped  are 
congruent  and  parallel. 

Ex.  1 .  The  edges  of  a  parallele- 
piped may  be  divided  into  three 
groups  of  four  parallel  edges  each. 

Ex.  2.  The  diagonals  of  a  parallelepiped  meet  in  a  point  which 
is  the  midpoint  of  each  diagonal. 

NOTE.  The  intersection  of  the  diagonals,  of  a  parallelepiped  is 
called  the  center  of  the  parallelepiped. 

Ex.  3.  Any  line  through  the  center  of  a  parallelepiped  and 
terminated  by  a  pair  of  opposite  faces  is  bisected  by  the  center  of 
the  parallelepiped. 

*  Let  the  pupil  construct  a  model  for  each  of  the  special  cases  mentioned.    These 
models  may  be  constructed  of  wood,  soap,  or  cardboard. 


FIG.  64 


54 


SOLID  GEOMETRY 


71.  A  parallelepiped  is  said  to  be  a  right  parallelepiped 
if  the  base  is  a  right  section. 

Ex.  1.  What  properties  has  a  right  parallelepiped  by  virtue 
of  the  fact  that  it  is  a  parallelepiped  and  at  the  same  time  a  right 
prism? 

Ex.  2.  Show  by  means  of  a  model  that  a  right  parallelepiped 
is  not  right  in  all  positions  in  which  it  may  be  placed. 

72.  A  right  parallelepiped  is  said  to  be  a  rectangular 
parallelepiped  if  its  base  is  a  rectangle  (Fig.  65). 

THEOREM  43.  If  a  parallelepiped  is  so  constructed  that 
each  of  its  three  edges  that  meet  at  a  common  vertex  is 
perpendicular  to  the  other  two,  the  parallelepiped  is  rec- 
tangular. 

H 

_^-* 


FIG.  65 

Analysis: 
I.  Let  each  of  the  edges  AB,  AD,  and  AE  be  _L  the 

other  two. 

II.  To  prove   that   the   parallelepiped   is  rectangular, 
prove  that 

(1)  the  base  ABCD  is  a  rectangle. 

(2)  the  parallelepiped  is 'a  right  parallelepiped. 
III.  To  prove  that  it  is  a  right  parallelepiped,  prove  the 

edges  AE,  BF,  CG,  and  DH  each  _L  the  base  ABCD. 

Ex.  1.  Each  face  of  a  rectangular  parallelepiped  is  a  rectangle. 

Ex.  2.  Show  that  a  rectangular  parallelepiped  is  a  right  paral- 
lelepiped in  all  positions  in  which  it  may  be  placed. 

Ex.  3.  The  diagonals  of  a  rectangular  parallelepiped  are  equal. 

Ex.  4.  Find  the  diagonal  of  a  rectangular  parallelepiped  if 
three  edges  that  meet  in  a  point  are  respectively  4,  8,  and  6  in. 
Find  the  diagonal  if  these  edges  are  o,  b,  and  c. 


POLYHEDRONS,  CYLINDERS,  AND  CONES         55 


73.  If  the  three  edges  of  a  rectangular  parallelepiped 
that  meet  at  one  vertex  are  equal,  the  rectangular  parallele- 
piped is  called  a  cube. 

THEOREM  44.    The  faces  of  a  cube  are  all  squares. 

Ex.  1.  Are  the  diagonals  of  a  cube  perpendicular  to  each 
other?    Why? 

Ex.  2.  If  the  edge  of  a  cube  is  E,  the  diagonal  is  E^S. 
Ex.  3.  One  edge  of  a  cube  is  5  in.    Find  the  area  of  a  section 
made  by  passing  a  plane  through  two  diagonally  opposite  edges. 

CYLINDERS 

DEFINITIONS 

74.  An  unlimited  closed  cylindrical 
surface  is  said  to  inclose  a  cylindrical 
space. 

A  solid  bounded  by  a  closed  cylin- 
drical surface  and  two  parallel  trans- 
verse sections  is  called  a  cylinder 
(Figs.  66  and  67).  It  is  a  portion  of 
a  cylindrical  space. 

The  transverse  sections  are  called 
the  bases  of  the  prism.  The  cylin- 
drical surface  and  its  elements  are 
called  respectively  the  lateral  surface 
and  the  elements  of  the  cylinder. 

The  perpendicular  distance  be- 
tween the  bases  of  the  cylinder  is- 
called  the  altitude  of  the  cylinder. 

A  cylinder  is  said  to  be  a  convex  cylin- 
der if  its  cylindrical  surface  is  convex. 

That  portion  of  a  cylinder  included 
between  one  base  and  a  transverse  section 
oblique  to  the  base  is  called  a  truncated 
cylinder  (Fig.  68).  FlG.  6g 


FIG.  66 


FIG.  67 


56  SOLID  GEOMETRY 

GENERAL  PROPERTIES  OF  CYLINDERS 

76.  THEOREM  45.    Two  parallel  transverse  sections  of  a 
closed  cylindrical  space  are  congruent. 


FIG.  69 

Hypothesis:     P  and  P'  are  two  parallel  transverse  sec- 
tions of  a  closed  cylindrical  space. 
Conclusion:    P  &  P'  . 
Analysis  and  construction: 
I.  To  prove  P  &  P',  prove  that  P  will  coincide  with  P' 

if  superposed. 

II.  Take  A,  B,  and  C  any  three  points  in  P.  Draw  the 
elements  AAf,  BB',  and  CC'  and  show  that  P  can 
be  placed  upon  P'  in  such  a  way  that  A,  B,  and  C 
fall  respectively  on  A',  B',  and  C'  and  at  the  same 
time  any  fourth  point  in  P  will  fall  on  a  point  in  P'. 
III.  To  prove  that  A,  B,  and  C  can  be  made  to  fall  on. 
A  ',.£',  and  C',  prove  AB  =  A'B',  AC  =  A'C',  and 


IV.  To  prove  that  a  fourth  point  in  P  falls  on  a  point  in 
P',  draw  the  element  XX'  and  prove 
and   ABAX=  /.B'A'X'. 
Outline  of  proof: 
I.  A  A'  ||  ££'. 

A'B'. 
AA'BB'isa  O. 


Similarly  AC  =  A'C'. 


POLYHEDRONS,  CYLINDERS,  AND  CONES          57 


Also  Z.BAC= 
II.  If  P  is  placed  on  P'  so  that  Z  B  AC  falls  on  Z.B'A'C, 
B  will  fall  on  B'  and  C  on  C  . 

III.  As  in  I,  AX=A'Xf  and  Z£AX  =  tB'A'X'. 

IV.  /.    if  P  is  placed  on  P'  so  that    Z.BAC  falls  on 

Z.B'A'C',  point  X  will  fall  on  point  X'. 
V.  In  the  same  way  every  point  on  P  will  fall  on  a  point 
on  P'  if  the  three  points  A,  B,  and  C  fall  on  the 
three  points  A',  B',  and  C'. 
Let  the  pupil  give  the  reasons. 

COR.    The  bases  of  a  closed  cylinder  are  congruent. 

76.  THEOREM  46.  If  a  plane  contains  one  element  of  a 
cylindrical  surface  and  one  other  point  of  that  surface,  then 
it  contains  the  element  through  that  point. 


FIG.  70 

Hypothesis:  C  is  a  cylindrical  surface.  P  is  a  plane 
containing  the  element  A  A'  and  the  point  K  of  the  cylin- 
drical surface. 

Conclusion:     P  contains  the  element  BB'  through  K. 

Analysis: 

I.  To  prove  that  P  contains  the  element  BB',  prove  that 
BB'  is  parallel  to  AA'. 

We  will  assume,  without  proof, 

As.  9.  If  C  is  a  closed  convex  cylindrical  surface,  the 
plane  P  (Th.  46)  will  intersect  the  cylindrical  surface  in  the 
two  elements  A  A'  and  BB'  and  in  no  other  points  (Fig.  70). 

5 


58  SOLID  GEOMETRY 

77.  COR.  Any  section  of  a  closed  convex  cylinder  made 
by  a  plane  containing  an  element  and  one  other  point  of 
the  cylindrical  surface  is  a  parallelogram. 


FIG.  .71 

Analysis: 

I.  To  prove  ABB' A'  a  O,  prove  each  side  parallel 
to  its  opposite. 

II.  To  prove  AB  \\  A'B' 

III.  To  prove  AA'  \\  BE',  prove  BB'  an  element  of  the 
cylinder  (Th.  46  and  As.  9). 

Ex.  1.  If  a  plane  contains  two  elements  of  a  convex  cylinder, 
the  section  of  the  cylinder  formed  by  this  plane  is  a  parallelogram. 

Ex.  2.  If  a  plane  contains  two  elements  of  a  cylinder,  it  is 
parallel  to  the  elements  of  the  cylinder. 

Ex.  3.  If  two  planes  each  containing  two  elements  of  a  cylinder 
intersect,  the  line  of  intersection  is  parallel  to  the  elements  of  the 
cylinder. 

Ex.  4.  Are  the  two  previous  exercises  true  if  the  planes  con- 
tain only  one  element  of  the  cylinder? 

RIGHT   CIRCULAR   CYLINDERS 

78.  Since  the  form  of  a  cylindrical  space  depends  upon 
the  form  of  its  right  section,  cylinders  may  be  named  from 
the  form  of  their  right  sections,  thus:  a  cylinder  is  called  a 
circular  cylinder  if  its  right  section  is  a  circle ;  it  is  called  an 
elliptical  cylinder  if  its  right  section  is  an  ellipse. 


POLYHEDRONS,  CYLINDERS,  AND  CONES         59 


FIG.  72 


NOTE.  Transverse  sections  of  circular  cylinders  that  are  not 
right  sections  are  ellipses.  This  fact  can  be  illustrated  by  means  of 
a  broomstick  cut  so  that  the  section  is  not  a  right  section.  An  ellip- 
tical cylinder  may  be  cut  by  a  plane  so  that  the  section  is  a  circle. 
In  higher  mathematics  other  kinds  of  cylinders  are  studied,  such  as 
hyperbolic  and  parabolic  cylinders. 

If  the  base  of  a  cylinder  is  a 
right  section,  the  cylinder  is  said 
to  be  a  right  cylinder  (Fig.  72). 
A  cylinder  that  is  not  a  right 
cylinder  is  called  an  oblique 
cylinder  (Fig.  73). 

A  right  cylinder  whose  base 
is  a  circle  is  called  a  right  circular 
cylinder. 

A  right  circular  cylinder 
may  be  generated  by  the 
revolution  of  a  rectangle 
about  one  side  as  an  axis. 
For  this  reason  a  right  cir- 
cular cylinder  is  sometimes 
called  a  cylinder  of  revolu- 
tion (Fig.  72). 

The  axis  about  which  the  rectangle  revolves  is  called  the 
axis  of  the  cylinder.  It  is  the  line  joining  the  centers  of  the 
bases.  The  sides  of  the  rectangle  perpendicular  to  the  axis 
generate  the  bases  of  the  cylinder.  The  side  parallel  to  the 
axis  generates  the  cylindrical  surface. 

Ex.  1.  What  is  a  cylindrical  surface  of  revolution?  How  may 
it  be  considered  a  locus? 

Ex.  2.  Every  section  of  a  right  convex  cylinder  made  by  a 
plane  containing  an  element  and  one  other  point  of  the  cylindrical 
surface  is  a  rectangle. 

Ex.  3.  Every  element  of  a  right  cylinder  is  equal  to  the  altitude m 

Ex.  4.  Every  section  of  a  right  circular  cylinder  made  by  a 
plane  parallel  to  the  base  is  a  circle. 


FIG.  73 


60  SOLID  GEOMETRY 

PYRAMIDS 
DEFINITIONS 

79.  A  closed  pyramidal  surface  is  said  to  inclose  a  pyram- 
idal space.     A  pyramidal  space  is  composed  of  two  parts 
called  nappes  which  correspond  to  the  two  nappes  of  the 
pyramidal  surface. 

A  solid  bounded  by  one  nappe  of  a  closed  pyramidal  sur- 
face and  a  transverse  section  is  called  a  pyramid  (see  Fig.  74) . 
It  is  a  portion  of  a  pyramidal  space. 

The  vertex  of  the  pyramidal  surface  is  called  the  vertex 
of  the  pyramid.  The  transverse  section  is  called  the  base 
of  the  pyramid.  The  pyramidal  surface  and  its  elements 
are  called  respectively  the  lateral  surface  and  the  lateral 
edges  of  the  pyramid. 

The  perpendicular  from  the  vertex  to  the  base  is  called 
the  altitude  of  the  pyramid. 

GENERAL  PROPERTIES  OF  PYRAMIDS 

80.  THEOREM  47.     If  a  pyramid  is  cut  by  a  plane  parallel 
to.  the  base, 

I.  The  edges  and  altitude  are  divided  proportionally. 
II.  The  section  is  a  polygon  similar  to  the  base. 


FIG.  74 

Suggestion  for  I.     Pass  a  plane  through-  0  parallel  to  P  and  P'. 
Use  Th.  30. 


POLYHEDRONS,  CYLINDERS,  AND  CONES          61 


Analysis  for  II: 
I.  To  prove  P~P',  prove 
a.    Z/1=ZA', 


',  ZC=ZC',  etc. 


AB      BC      CD  _ 
''  A'B'~B'C'~C'D'~ 
II.  To  prove  /.A=  /.A',  prove 

AB      BC  AB      OB      BC 

III.  To  prove  -— ,  =  -— -/f  prove  -77=  =  -^7  =  -=™ 


A'E',  AB  \\  A'B'. 


COR.  I.  If  a  pyramid  is  cut  by  a  plane  parallel  to  the 
base,  the  ratio  of  the  area  of  the  section  to  the  area  of  the 
base  equals  the  ratio  of  the  squares  of  their  distances  from 
the  vertex. 

areaP     "OX*  areaP       ~AB2 

Analysis:  1  o  prove  -    —  =-f  =        n  ,  prove  -  =r.  =  - 
areaP'  >  areaP'        T/2' 


DA 

and  -  -2  (Plane  Geometry, Th.  127)  (Fig.  74) . 

A'£'       OA'       OX' 

COR.  II.  If  two  pyramids  have  equal  altitudes  and 
equal  bases,  sections  parallel  to  the  bases  at  equal  distances 
from  the  vertices  are  equal. 


FIG.  75 


a     ,     .       >„  ,  ,  areas'     OK1 

Analysis:     To  prove  area  s'  =  area  r  ,  prove  --  =  -  ^ 

area  5          - 


and  -  (see  P/aw^  Geometry,  Th.  92)  . 

arear  2 


62 


SOLID  GEOMETRY 


Ex.  1.  At  what  distance  from  the  vertex  of  a  pyramid  must 
a  plane  be  passed  parallel  to  the  base  so  that  the  area  of  the  sec- 
tion may  be  one-half  the  area  of  the  base? 

Ex.  2.  If  a  plane  is  passed  parallel  to  the  base  of  a  pyramid 
and  through  the  mid-point  of  the  altitude,  what  is  the  ratio  of  the 
area  of  the  section  to  the  area  of  the  base? 

Ex.  3.  It  is  stated  in  physics  that  the  intensity  of  illumination 
received  on  a  screen  is  inversely  proportional  to  the  square  of  the 
distance  of  the  screen  from  the  source  of  light.  Show  that  this  is 
an  application  of  Cor.  I,  Th.  47. 

REGULAR  PYRAMIDS 

81.  A  pyramid  is  said  to  be  regular  if  its  base  is  a  regular 
polygon  and  the  altitude  passes  through  the  center  of  the 
base  (Fig.  76). 

THEOREM  48.  The  lateral  edges  of  a  regular  pyramid 
are  equal. 

THEOREM  49.  The  lateral  faces  of  a  regular  pyramid  are 
congruent  isosceles  triangles. 

THEOREM  50.  The  altitudes  of  all  of  the  lateral  faces 
of  a  regular  pyramid  are  equal. 

The  altitude  of  any  one  of  the  lateral  faces  of  a  regular 
pyramid  is  called  the  slant  height  of  the  pyramid. 

Ex.  1.  Any  section  of  a  regular  square  pyramid  made  by  a 
plane  containing  the  altitude  is  an  isosceles  triangle. 

Ex.  2.  The  altitudes  of 
the  face  triangles  of  a  regular 
pyramid  meet  the  base  at  the 
points  of  tangency  of  the  circle 
inscribed  in  the  base  (Fig.  76). 

Suggestion.     Join  Ot  the  foot 
of  the  altitude  of  the  pyramid, 
with  Y,  the  point  of  tangency  of 
AB.     Join   YX  and  prove    YX  <- 
the  altitude  of  A  ABX.  FIG.  70 


POLYHEDRONS,  CYLINDERS,  AND  CONES         63 


FIG.  78 


FRUSTUMS   OF   PYRAMIDS 

82.  The  portion  of  a  pyramid 
included  between  the  base  and  a 
section  oblique  to  the  base  cut- 
ting all  of  the  lateral  edges  is 
called  a  truncated  pyramid 
(Fig.  77). 

The  portion  of  a  pyramid  in- 
cluded between  the  base  and  a 
section   parallel   to   the   base 
and  cutting  all  of  the  lateral 
edges  is  called  a  frustum  of 
a  pyramid  (Fig.  78). 

The  altitude  of  a  frustum 
of  a  pyramid  is  the  perpen- 
dicular distance  between  the    - 
bases  (YY',  Fig.  78). 

THEOREM  51.  The  lateral  faces  of  a  frustum  of  a  pyra- 
mid are  trapezoids. 

THEOREM  52.  The  lateral  faces  of  a  frustum  of  a  regular 
pyramid  are  congruent  isosceles  trapezoids. 

THEOREM  53.  The  altitudes  of  the  faces  of  a  frustum 
of  a  regular  pyramid  are  equal. 

The  altitude  of  any  face  of  a  frustum  of  a  regular  pyra- 
mid is  called  the  slant  height  of  the  frustum. 

Ex.  1.  The  sum  of  the  segments  connecting  the  mid -points  of 
the  lateral  edges  of  a  frustum  of  a  pyramid  is  one-half  the  sum  of 
the  perimeters  of  the  bases. 

Ex.  2.  Prove  that  the  medians  re- 
ferred to  in  the  previous  exercise  are 
co-planar. 

Suggestion.  Let  X  be  the  mid-point 
of  A  A' '.  Through  X  pass  a  plane  parallel 
to  ABCDE.  Prove  that  the  medians  of 
the  several  lateral  faces  lie  in  this  plane 
(Fig.  791.  FIG.  79 


A' 


04  SOLID  GEOMETRY 

CONES 

DEFINITIONS 

83.  A  closed  conical  surface  is  said  to  inclose  a  conical 
space.  A  conical  space  is  composed  of  two  parts  called 
nappes,  which  correspond  to  the  two  nappes  of  the  conical 
surface. 

A  solid  bounded  by  one  nappe  of  a  closed  conical  surface 
and  a  transverse  section  is  called  a  cone  (Fig.  80).  It  is  a 
portion  of  ,a  conical  space. 

The  vertex  of  the  conical  surface  is  called  the  vertex  of 
the  cone.  The  transverse  section  is  called  the  base  of  the 
cone.  The  conical  surface  and  its  elements  are  called 
respectively  the  lateral  surface  and  the  elements  of  the  cone. 

The  perpendicular  from  the  vertex  to  the  base  is  called 
the  altitude  of  the  cone. 


GENERAL  PROPERTIES  OF  CONES 

84.  THEOREM  54.    If  the  base  of  a  cone  is  a  circle,  every 
section  parallel  to  the  base  is  also  a  circle. 


FIG.  80 

Analysis  and  construction: 
I.    To  prove  pf  a  circle,  prove  that  all  points  on  pr  are 

equally  distant  from  a  point  within  p' '. 
II.     .*.  choose  C1  and  Dr  any  two  points  on  p',  and  prove 

C*  and  D'  equally  distant  from  a  point  (X')  within  p'. 


POLYHEDRONS,  CYLINDERS,  AND  CONES 


65 


III.  To  find  X',  join  0  and  X,  the  center  of  Op.     Let 

OX  cut  p'  at  X'. 

IV.  /.  join  C'X'  and  £>'X'  and  prove  -C'X'^D'X'. 

V.  To  prove  C'X'  =  D'X',  draw  the  elements  OC'  and 
OD'\  let  the  planes  COX'  and  D'OJT  cut  £  in 
CX  and  £>X  respectively,  and  prove 
CJC    OJC        '  D'X'  =  OX' 
CX  'OX'*       DX      OX' 

COR.  I.  If  the  base  of  a  cone  is  a  circle,  the  area  of  a 
section  parallel  to  the  base  is  to  the  area  of  the  base  as  the 
square  of  the  distance  of  the  section  from  the  vertex  is  to 
the  square  of  the  altitude. 


FIG.  81 


FIG.  82 


area  p'     OK' 

Analysis:     To  prove  --  —  =        ,  ,  prove 
area  p      QK* 


=      =  = 

area/?       r2     QX2      OK2 
COR.  II.     If  two  cones  have  equal  circular  bases  and 
equal  altitudes,  areas  of  sections  at  equal  distances  from 
the  vertices  are  equal. 


Analysis:    To  prove  p'  =  s',  prove 


area  s'     01" 


area  p 


and 


area  5      VY     (Fig.  82). 

Exercise.     Investigate  Th.  54  and  its  two  corollaries  if  the  plane 
fo  ming  section  p  cuts  the  other  nappe  of  the  conical  surface. 


06  SOLID  GEOMETRY 

85.  THEOREM  55.  If  a  plane  contains  one  element  of  a 
conical  surface  and  one  other  point  of  that  surface,  then  it 
contains  the  element  through  that  point. 


FIG.  83 

Hypothesis:  OAB  is  a  conical  surface  cut  by  the  plane  P 
which  contains  the  element  OC  and  the  point  K  of  the  conical 
surface. 

Conclusion:  The  plane  P  contains  the  element  OD 
through  K. 

Analysis:  To  prove  that  P  contains  the  element  OD 
through  K,  show  that 

a.  Points  K  and  0  are  both  in  P. 

b.  Points  K  and  0  determine  an  element. 

If  the  conical  surface  of  a  cone  is  convex,  the  cone  is  said 
to  be  a  convex  cone.  Unless  otherwise  stated,  all  cones 
referred  to  will  be  considered  convex  cones. 

We  will  assume,  without  proof, 

As.  10.  If  C  is  a  closed  convex  conical  surface,  the 
plane  P  (Th.  55)  will  intersect  the  conical  surface  in  the  two 
elements  CO  and  DO  and  in  no  other  point  (Fig.  83). 

COR.  If  a  plane  contains  one  element  of  a  conical  sur- 
face and  one  other  point  of  that  surface,  the  section  of  the 
cone  made  by  the  plane  is  a  triangle. 

Exercise.  Any  section  of  a  cone  made  by  a  plane  that  passes 
through  the  vertex  and  intersects  the  base  is  a  triangle. 


POLYHEDRONS,  CYLINDERS,  AND  CONES          67 
RIGHT   CIRCULAR   CONES 

86.  If  a  cone  has  a  circular  section  such  that  the  line  from 
the  vertex  to  the  center  of  the  section  is  perpendicular  to 
the  plane  of  that  section,  the  cone  is  called  a  circular  cone. 

If  the  base  of  a  cone  is  a  circle,  and  if  the  line  from  the 
vertex  to  the  center  of  the  base  is  perpendicular  to  the  base, 
the  cone  is  called  a  right  circular  cone. 

A  circular  cone  that  is  not  a  right  circular  cone  is  called 
an  oblique  circular  cone. 

THEOREM  56.  All  elements  of  a  right  circular  cone  are 
equal. 

An  element  of  a  right  circular  cone  is  called  its  slant 
height. 

A  right  circular  cone  may  be  gener- 
ated by  the  revolution  of  a  right  tri- 
angle about  one  leg  as  an  axis.  For 
this  reason  a  right  circular  cone  is 
sometimes  called  a  cone  of  revolution  ______ __ 

(Fig.  84).  FIG.  84 

The  axis  about  which  the  triangle  revolves  is  called  the 
axis  of  the  cone.  The  other  leg  generates  the  base  of  the 
cone.  The  hypotenuse  generates  the  conical  surface. 

The  axis  of  a  right  circular  cone  is,  therefore,  the  segment 
from  the  vertex  to  the  center  of  the  base ;  it  is  perpendicular 
to  the  base  and  is  the  altitude  of  the  cone. 

Ex.  1.  What  is  a  conical  surface  of  revolution?  Why  may  it 
be  considered  a  locus? 

Ex.  2.  A  section  of  a  right  circular  cone  made  by  a  plane 
containing  the  vertex  and  intersecting  the  base  is  an  isosceles 
triangle. 

Ex.  3.  What  kind  of  a  triangle  is  the  section  referred  to  in  the 
previous  exercise  if  the  cone  is  an  oblique  circular  cone? 


68 


SOLID  GEOMETRY 


Ex.  4.  Can  a  section  of  an  oblique  circular  cone  made  by  a 
plane  which  contains  the  vertex  and  intersects  the  base  be  an 
isosceles  triangle?  How?  Give  proof. 

NOTE.     The  various  forms  of  the  sections  of  a  right  circular  cone 
are  interesting.     The  complete  study  of  them  is  beyond  this  book. 

For  each  case  shown  in  the  figure  the  construction  of  the  cutting 
plane  and  the  form  of  the  section  are  given  below: 

No.  1.  Contains  the  vertex  and  intersects  the  base.     A  triangle. 

No.  2.  Parallel  to  the  base.     A  circle. 

No.  3.  Cuts  all  of  the  elements  of  the  same  nappe,  but  not  parallel 
to  the  base.  An  ellipse. 

No.  4.  Parallel  to  an  element.     A  parabola. 

No.  5.  Perpendicular  to  the  base.  The  plane  cuts  both  nappes. 
The  section  has  two  parts.  A  hyperbola. 


No.  1 


No.  2 


No.  3 
FIG.  85 


No.  4 


No.  5 


FRUSTUMS    OF   CONES 

87.  The  portion  of  a  cone  included 
between  the  base  and  a  transverse  sec- 
tion oblique  to  the  base  is  called  a 
truncated  cone  (Fig.  86). 


The  portion  of  a  cone  included 
between  the  base  and  a  transverse 
section  parallel  to  the  base  is  called 
the  frustum  of  a  cone  (Fig.  87). 


FIG.  86 


FIG.  87 


POLYHEDRONS.,  CYLINDERS,  AND  CONES          69 

THEOREM  57.  The  elements  of  a  frustum  of  a  right 
circular  cone  are  equal. 

An  element  of  a  frustum  of  a  right  circular  cone  is  called 
the  slant  height  of  the  frustum. 

Ex.  1.  Any  section  of  a  frustum  of  a  cone  made  by  a  plane 
containing  an  element  and  one  other  point  in  the  conical  surface 
is  a  trapezoid. 

Ex.  2.  Show  that  if  the  frustum  referred  to  in  the  previous 
exercise  had  been  the  frustum  of  a  right  circular  cone  the  section 
would   have    been    an    isosceles 
trapezoid.  /CZZHX 

Ex.  3.  The  perimeter  of  a  sec- 
tion of  a  frustum  of  a  cone  made 
by  a  plane  parallel  to  the  bases 
and  midway  between  them  is 
equal  to  one-half  the  sum  of 

the  perimeters  of  the  two  bases 

(Fig.  88).     •  FIG.  88 

A  frustum  of  a  right  circular  cone  may  be  generated  by 
the  revolution  of  an  isosceles  trapezoid  about  the  segment 
joining  the  mid-points  of  the  bases  as  an  axis.  For  this 
reason  a  frustum  of  a  right  circular  cone  may  be  called  a 
frustum  of  revolution. 

The  axis  about  which  the  trapezoid  revolves  is  called  the 
axis  of  the  frustum.  The  bases  of  the  trapezoid  generate 
the  bases  of  the  frustum. 

REGULAR  POLYHEDRONS 

88.  A  polyhedron  is  said  to  be  regular  if  its  faces  are 
congruent  regular  polygons  and  if  its  polyhedral  angles  are 
congruent. 

Polyhedrons  may  be  named  from  the  number  of  faces: 
A  polyhedron  of  4  faces  is  called  a  tetrahedron;  of  6  faces, 
a  hexahedron;  of  8  faces,  an  octahedron;  of  12  faces,  a 
dodecahedron;  of  20  faces,  an  icosahedron. 


70  SOLID  GEOMETRY 

89.  THEOREM  58.  Not  more  than  five  regular  poly- 
hedrons are  possible. 

Outline  of  proof: 

A.  In  general. 

I.  The  faces  of  the  regular  polyhedrons  must  be  equi- 
lateral triangles,  squares,  regular  pentagons,  or  some  other 
regular  polygons. 

II.  At  least  three  faces  must  meet  at  each  vertex  to  form 
a  polyhedral  angle. 

III.  The  sum  of  the  face  angles  of  each  polyhedral  angle 
must  be  less  than  360°. 

B.  When  the  faces  are  equilateral  triangles. 

How  many  degrees  in  each  angle  of  an  equilateral  triangle  ? 

What  would  be  the  sum  of  the  face  angles  of  a  polyhedral 
angle  if  3  equilateral  triangles  met  at  the  vertex  ?  If  4  such 
triangles  met  at  a  vertex?  If  5?  If  6  or  more?  Tabulate 
the  results. 

How  many  regular  polyhedrons  could  be  formed  whose 
faces  are  equilateral  triangles?  How  many  triangles  meet 
at  each  vertex  ? 

C.  When  the  faces  are  squares. 

What  would  be  the  sum  of  the  face  angles  of  a  polyhedral 
angle  if  3  squares  met  at  the  vertex?  If  4  squares  met  at 
the  vertex  ?-  If  more  than  4  squares  were  used  ? 

How  many  regular  polyhedrons  could  be  formed  whose 
faces  are  squares?  How  many  squares  would  meet  at  each 
vertex  ? 

D.  When  the 'faces  are  regular  pentagons. 

In  the  same  way,  how  many  regular  polyhedrons  could 
be  formed  whose  faces  would  be  regular  pentagons?  How 
many  pentagons  would  meet  at  each  vertex  ? 

E.  When  the  faces  are  other  regular  polygons. 

Show  that  in  this  case  no  regular  polyhedron  could  be 
formed. 


POLYHEDRONS,   CYLINDERS,  AND   CONES         71 

Exercise.     Tabulate  the  results  of  your  study  of  Th.  58,  show- 
ing in  your  table 

(1)  the  kinds  of  polygons  used  as  faces 

(2)  the  number  of  faces  that  meet  at  each  vertex 

(3)  the  sum  of  the  face  angles  at  each  vertex 

(4)  the  total  number  of  faces 

(5)  the  name  of  the  polyhedron 

90.  There  are  five  regular  polyhedrons.     This  fact  may 
be  verified  by  models  made  by  the  pupil. 


FIG.  89 


Fig.  89  shows  these  polyhedrons,  and  Fig.  00  shows  the 
patterns  for  making  them. 


No.  5 


No.  4 


FIG.  90 


NOTE.  When  making  the  models,  cut  out  the  patterns,  cutting 
only  halfway  through  along  the  dotted  lines.  Fold  along  the  dotted 
lines  and  fasten  the  edges  together  with  gummed  paper. 


72 


SOLID    GEOMETRY 


91.  NOTE.    Star  polyhedrons.   Fig.  91  shows  two  star  polyhedrons. 

To  make  No.  1,  construct  a  regular  dodecahedron  and  paste  a  pyra- 
mid upon  each  face.  To  construct  the  pyramids,  draw  a  regular 
pentagon  congruent  with  one  of  the  faces  of  the  dodecahedron ;  extend 
the  sides  of  the  pentagon,  forming  a  pentagram-star;  turn  up  the  tri- 
angles to  form  the  pyramid  whose  base  is  the  regular  pentagon. 


No.  1 


No.  2 


FIG.  91 


To  make  No.  2,  paste  on  each  face  of  the  regular  icosahedron 
a  regular  tetrahedron  one  face  of  which  is  congruent  with  the  faces 
of  the  icosahedron. 

The  strings  joining  the  vertices  of  these  pyramids  form  what  has 
been  called  the  case  of  the  star  polyhedron.  The  case  of  No.  1  is  the 
regular  icosahedron.  The  case  of  No.  2  is  the  regular  dodecahedron. 
The  regular  polyhedron  used  as  the  basis  has  been  called  the  core  of 
the  star  polyhedron. 

SUPPLEMENTARY  EXERCISES 

92.  1.  Name  a  solid  which  has  for  faces  (1)  four  equilateral 
triangles,  (2)  two  triangles  and  three  parallelograms,  (3)  two 
triangles  and  three  rectangles,  (4)  one  square  and  four  triangles, 
(5)  six  parallelograms,  (6)  two  parallelograms  and  four  rectangles, 
(7)  six  rectangles,  (8)  eight  equilateral  triangles. 

2.  Construct  a  parallelepiped  that  has  its  edges  on  three  given 
skew  lines. 

3.  Given  three  segments  equal  to  the  three  edges  of  a  rectangu- 
lar parallelepiped  th'at  meet  in  one  vertex.     Construct  a  segment 
equal  to  the  diagonal. 


POLYHEDRONS,    CYLINDERS,   AND   CONES         73 


4.  What  is  the  form  of  a  section  of  a  prism  made  by  a  plane 
which  is  (1)  parallel  to  a  lateral  edge  and  intersects  the  base;  (2) 
parallel  to  one  lateral  face  and  intersects  the  base?     Give  proof. 

5.  What  is  the  answer  to  Ex.  4  if  the  prism  is  a  right  prism? 

6.  What  is  the  section  of  a  parallelepiped  made  by  a  plane 
located  as  given  below:  (1)  if  the  parallelepiped  is  oblique;  (2)  if 
it  is  right;   (3)  if  it  is  rectangular: 

a.  Passes  through  two  diagonally  opposite  edges? 

b.  Cuts  four  parallel  edges? 

c.  Is  perpendicular  to  a  face  and  cuts  four  parallel  edges? 

d.  Is  perpendicular  to  an  edge  and  cuts  four  parallel  edges? 

7.  How  can  you  cut  a  cube  so  that  the  section  will  be  (1)  a 
parallelogram,  (2)   a  rectangle,   (3)   a  square,   (4)  an  equilateral 
triangle,  (5)  an  isosceles  triangle, 

(6)  a  regular  hexagon? 

Analysis  for  6  (Fig.  92): 

I.  To     prove     UVWXYZ    a 
regular  hexagon,  prove 

(1)  C7,  VtW,X,  Y,Z  are  co- 

planar. 

(2)  UVWXYZ  is  equilateral. 

(3)  UVWXYZ    can   be 'in- 

scribed in  a  circle. 

IT.  To  prove  I,  prove  VY,  UX, 
and  WZ  each  _L  FD  at  0 
(the  mid-point  of  FD). 
III.  .'.  show  (1)  that  VY  lies  in 
plane  EBCH;  (2)  that  O 
and  F  are  equally  distant 
from  the  extremities  of  VY. 


FIG.  93 


8.  The  section  of  a  tetrahedron 
made  by  a   plane   parallel   to   two 
non-intersecting  edges  is  a  parallelogram  (Fig.  93). 

9.  Find  the  sum  of  the  plane  angles  of  the  dihedral  angles  whose 
edges  are  the  lateral  edges  of  a  triangular  prism. 

Suggestion.     Construct  a  right  section  of  the  prism. 
6 


74 


SOLID  GEOMETRY 


FIG.  94 


10.  What  is  the  sum  of  the  plane  angles  referred  to  in  Ex.  9 
if  the  prism  is   quadrangular  instead  of  triangular?     Can   the 
exercise  be  solved  for  any  other  kind  of  prisms? 

11.  If  the  bases  of  a  cylinder  are  circles,  a 
line  parallel  to  an  element  of  a  cylinder  contain- 
ing the   center  of  one   base   contains   also   the 
center  of  the  other  base. 

Analysis  and  construction  (Fig.  94): 
I.  To  prove  0'  the  center  of  O  A'B',  prove  that 
any  two  points   in  O   A'B'  are  equally 
distant  from  0'. 

11.  That  is,  if  A'  and  C  are  any  two  points  in  0  A'B',  prove 

0'A'  =  0'C'. 

III.    .'.  draw  the  elements  through  A'  and  C'  meeting  base  AB  in 
A  and  C  respectively,  and  prove  A'0'  =  AO  =  OC  =  0'C. 

If  the  bases  of  a  cylinder  are  circles,  the  line  joining  the 
centers  of  the  bases  is  called  the  axis  of  the  cylinder. 

12.  If  the  bases  of  a  cylinder  are  circles,  the  axis  of  the  cylinder 
is  parallel  to  the  elements. 

Suggestion.     Use  Ex.   11   and  prove  that  the  axis 
coincides  with  a  line  that  is  parallel  to  the  elements. 

13.  In  every  cylinder  whose  bakes  are  circles 
there  is  one  set  of  parallel  planes  that  cut  the 
cylinder  in  sections  that  are  rectangles. 

Suggestion  (Fig.  95).     Let  00'  be  the  axis  of  the 
cylinder.     Construct  CD,  the  projection  of  00'  upon 
the   lower   base.     Draw    AB  J_  CD    at    0.     Pass   a          FIG 
plane  through  AB  and  the  element  through  A. 

To  prove   A'A    L  AB,   prove  OOr  _L   AB,  and  OO'  \\AA'. 

14.  How  many  vertices  and  how  many  edges  has  a  tetrahedron? 
A  regular  octahedron?    A  prism  whose  base  has  5  sides?    6  sides? 
n  sides? 

regular  poly- 


15.  Verify  the  following   formula  for  the 
hedrons: 


POLYHEDRONS,  CYLINDERS,  AND  CONES          75 


where  F,  F,  and  E  are  the  number  of  faces,  vertices,  and  edges 
respectively. 

NOTE.     The  formula  above  is  known  as  Euler's  Theorem.     It  is 
true  for  all  convex  polyhedrons.     A  proof  is  too  difficult  for  this  book. 

16.  Construct  a  regular  tetrahedron 
(Fig.  96). 

Suggestion.  Construct  the  equilateral 
triangle  ABC.  Let  O  be  the  center  of 
the  circle  circumscribed  about  A  ABC. 
Construct  OX  J_  plane  of  ABC  at  0. 
Find  X  so  that  XB=AB. 


FIG.  96 


1 7.  Construct  a  regular  octahedron  (Fig.  97) . 

Suggestion.  Construct  the  square  A  BCD. 
Let  0  be  the  intersection  of  the  diagonals  AC 
and  BD.  Construct  EF  _L  A  BCD  at  0.  Take 
OE  =  OF  =  OA.  Join  EA,  EB,  EC,  etc.  A 

18.  The  altitude  of  a  regular  tetrahedron 
meets   the    base    at    the    intersection   of   the 
medians. 


E  i7- 


19.  If  an  edge  of  a  regular  tetrahedron  is  E,  the  altitude  is  -5  V6. 

o 

20.  The  segments  joining  in  order  the  mid-points  of  two  pairs  of 
opposite  sides  of  a  regular    tetra- 
hedron form  a  square  (Fig.  98). 

21.  The  mid-points  of  the  edges 
of  a  regular  tetrahedron  are  the 
vertices  of  a  regular  octahedron. 

22.  Inscribe  a  regular  octahe- 
dron in  a  cube.  FIG.  98 


CHAPTER  III 

THE  SPHERE 

INTRODUCTORY 

DEFINITIONS 

93.  A  spherical  surface  has  been  defined  as  the  locus  of 
points  in  space  at  a  given  distance  from  a  given  point. 

The  given  point  is  called  the  center. 

The  given  distance  is  called  the  radius. 

A  sphere  is  a  solid  bounded  by  a  spherical  surface. 

The  center  and  the  radius  of  the  spherical  surface  are 
called  respectively  the  center  and  the  radius  of  the  sphere. 

A  segment  through  the  center  of  the  sphere  and  termi- 
nating in  the  spherical  surface  is  called  a  diameter  of  the 
sphere. 

ASSUMPTIONS   CONCERNING   SPHERES 

94.  We  will  assume  the  following: 

As.  11.  Spheres  with  equal  radii  or  with  equal  diameters 
are  congruent. 

As.  12.  All  radii  of  the  same  sphere  or  of  congruent 
spheres  are  equal. 

As.  13.  All  diameters  of  the  same  or  of  congruent  spheres 
are  equal. 

As.  14.  A  diameter  of  a  sphere  is  twice  the  radius  of  that 
sphere. 

As.  15.  A  sphere  is  located  definitely  if  its  center  and  its 
radius  are  known. 

As.  16.  A  segment  joining  a  point  within  a  sphere  and  the 
center  is  shorter  than  the  radius. 

76 


THE  SPHERE  77 

As.  17.  If  a  segment  that  has  one  end  at  the  center  of 
the  sphere  is  shorter  than  the  radius,  it  lies  wholly  within  the 
sphere. 

As.  18.  A  segment  joining  a  point  without  a  sphere  and 
the  center  is  longer  than  the  radius. 

As.  19.  If  a  segment  that  has  one  end  at  the  center  of 
the  sphere  is  longer  than  the  radius,  it  extends  without  the 
sphere  and  cuts  the  sphere  but  once. 

As.  20.  If  a  straight  line  passes  through  a  point  within 
a  sphere,  it  intersects  the  sphere  in  two  and  only  two  points. 

As.  21.  If  a  straight  line  passes  through  a  point  without 
the  sphere,  it  may  (1)  have  no  point  in  common  with  the 
sphere,  (2)  have  one  point  in  common  with  the  sphere,  or 
(3)  it  may  intersect  the  sphere  in  two  and  only  two  points. 

As.  22.  A  plane  through  a  point  within  a  sphere  inter- 
sects the  sphere  in  a  closed  curve. 

As.  23.  A  plane  through  a  point  without  a  sphere  may 
(1)  have  no  point  in  common  with  the  sphere,  (2)  have  one 
point  in  common  with  the  sphere,  or  (3)  may  intersect  the 
sphere  in  a  closed  curve. 

96.  The  following  preliminary  theorem  is  evident: 

THEOREM  59.  If  a  plane  is  passed  through  the  center 
of  the  sphere,  the  intersection  of  the  plane  and  the  spherical 
surface  is  a  circle. 

TANGENTS  TO   SPHERES 

FUNDAMENTAL   THEOREM 

96.  A  line  or  a  plane  that  has  one  point  in  common  with 
the  sphere,  but1  does  not  intersect  it,  is  said  to  be  tangent 
to  the  sphere. 

The  point  which  the  tangent  line  or  plane  has  in  common 
with  the  sphere  is  called  the  point  of  contact  or  the  point  of 
tangency. 


78 


SOLID  GEOMETRY 


97.  THEOREM  60.  A  plane  that  is  perpendicular  to  a 
radius  of  a  sphere  at  its  outer  extremity  is  tangent  to  the 
sphere. 


FIG.  99 

Hypothesis:     0  is  any  sphere  with  center  O.     OA  is  any 
radius  of  sphere  O.     Plane  M  is  perpendicular  to  OA  at  A. 
Conclusion:     Plane  M  is  tangent  to  sphere  0. 
Analysis  and  construction: 

I.  To  prove  plane  M  tangent  to  sphere  0  at  A,  prove 
that  all  points  in  M,  except  A,  lie  outside  the 
sphere. 
II.   /.  take  B  any  point  in  M  except  A,  join  B  and  0, 

and  prove  that  B  lies  outside  the  sphere. 
III.   /.  prove  OB  >  OA.    (See  Th.  14  and  As.  19.) 

THEOREM  61.  A  plane  that  is  tangent  to  a  sphere  is  per- 
pendicular to  the  radius  of  the  sphere  at  the  point  of  contact. 

Analysis  and  construction:  Use  an  indirect  proof.  If  OA  is 
not  J_  M,  suppose  some  other  line  as  OB  _L  M  and  show  that 
the  supposition  that  OB  _L  M  contradicts  the  hypothesis. 

,  Outline  of  proof: 

I.  If  OB  _L  M,  OB<OA. 
II.  If  OB  <  OA,  point  B  is  within  the  sphere. 
III.  If  B  is  within  the  sphere,  M  is  not  tangent  to  the 

sphere  (As.  22). 

IV.  But  M  is  given  tangent  to  the  sphere. 
V.    .'.  OA  is  _L  M. 


THE  SPHERE  79 

Ex.  1.  Are  Ths.  60  and  61  true  for  straight  lines  instead  of 
planes?  Give  proof. 

Ex.  2.  If  a  plane  is  tangent  to  a  sphere,  every  line  in  the  plane 
drawn  through  the  point  of  contact  is  tangent  to  the  sphere'. 

Ex.  3.  All  lines  tangent  to  a  sphere  at  a  given  point  on  the 
sphere  lie  in  the  plane  tangent  to  the  sphere  at  that  point. 

Ex.  4.  If  two  lines  are  tangent  to  a  sphere  at  the  same  point, 
the  plane  of  these  lines  is  tangent  to  the  sphere  at  that  point. 

Ex.  5.  Show  how  to  construct  a  line  that  shall  be  tangent  to  a 
given  sphere  and  shall  contain  a  given  point.  Discuss  three  cases. 

Suggestion.  Reduce  to  a  plane  geometry  construction  by  passing 
any  plane  through  the  given  point  and  the  center  of  the  sphere. 

Ex.  6.  Answer  Ex.  5  for  a  plane  tangent  to  the  sphere  instead 
of  for  a  line  tangent  to  the  sphere. 

Ex.  7.  All  segments  tangent  to  a  sphere  from  a  given  point 
without  the  sphere  are  equal. 

CIRCUMSCRIBED    AND   INSCRIBED   SPHERES 

98.  If  all  the  faces  of  a  polyhedron  are  tangent  to  a 
sphere,  the  polyhedron  is  said  to  be  circumscribed  about  the 
sphere  and  the  sphere  is  said  to  be  inscribed  in  the  poly- 
hedron. 

If  all  the  vertices  of  a  polyhedron  lie  in  a  given  spherical 
surface,  the  polyhedron  is  said  to  be  inscribed  in  the  sphere 
and  the  sphere  is  said  to  be  circumscribed  about  the  poly- 
hedron. 

DETERMINATION   OF   SPHERES 

99.  The  following  exercise  is  preliminary  and  may  be 
quoted  as  a  theorem  in  proving  Th.  62. 

Exercise.  Lines  that  are  perpendicular  to  two  intersecting 
planes  cannot  be  parallel. 

Analysis:     Use  indirect  proof. 

I.  Show  that  if  a\\b,M  would  be  ||  N.  b 

II.  To  show  that  M  would  be  ||  N,  show  M  and  .V 

would  both  be  X  a.  FIG.  100 


80  SOLID  GEOMETRY 

THEOREM  62.     One  and  only  one  spherical  surface  can 
be  passed  through  four  given  points  that  are  not  co-planar. 


FIG.  101 

Hypothesis:     ABCD  is  any  tetrahedron. 
Conclusion: 

a.  A  sphere  can  be  passed  through  points  A,  B,  C,  and  D. 

b.  Only  one  sphere  can  be  passed  through  A,  B,  C,  and 

D. 

Analysis  in  general  for  a: 
I.  It  is  necessary  to  prove  that  there  is  a  point  equally 

distant  from  A,  B,C,  and  D. 
II.    .*.  find  the  locus  of  points  equally  distant  from  A, 

B,  and  C  and  the  locus  of  points  equally  distant 

from  A,  B,  and  D.     Prove  that  these  loci  intersect. 
Construction: 
I.  To  find  the  locus  of  points  equally  distant  from  A, 

B,  and  D,  find  the  perpendicular  to  the  plane  of  A, 

B}  and  D  at  the  center  of  the  circle  circumscribing 

AABD.     Let  this  locus  be  XV. 
II.  In  the.  same  way  find  the  locus  of  points  equally 

distant  from  A,  B,  and  C.     Let  this  locus  be  ZW. 

'NOTE.  To  prove  XV  and  ZW  intersect,  prove  that  they  lie  in  the 
same  plane  and  are  not  parallel.  .*.  prove  XYand  ZW  are  in  a  plane 
_L  AB  at  its  mid-point. 

Analysis  for  b:    Show  that  there  is  only  one  point  equally 
distant  from  points  A,  B,  C,  and  D. 


THE   SPHERE 
CIRCLES   OF   SPHERES 


81 


FUNDAMENTAL   THEOREM 

100.  THEOREM  63.    Every  section  of  a  spherical  surface 
made  by  a  plane  is  a  circle. 


FIG.  102 

Hypothesis:     0  is  the  center  of  the  given  sphere  cut  by 
plane  M  in  section  A  BCD. 

Conclusion:     The  section  A  BCD  is  a  circle. 
Analysis  and  construction: 

I.  To  prove  ABCD  a  circle,  prove  that  every  point  in 
A  BCD  is  equally  distant  from  some  point  within. 
.  draw  a  perpendicular  from  O  to  plane  M  meeting 
plane  M  at  point  X.     Join  X  with  B  and  C,  any 
two  points  in  ABCD,  and  prove  BX  =  CX. 
join  BO  and  CO  and  prove  ABXO  =  ACXO. 
Since  we  have  seen  in  Th.  59  that  a  plane  passing  through 


II 


III. 

NOTE. 


the  center  of  the  sphere  is  a  circle,  it  is  only  necessary  in  Th.  63  to  prove 
that  the  section  of  the  spherical  surface  is  a  circle  if  the  plane  does 
not  pass  through  the  center. 

DEFINITIONS 

101.  A  section  of  a  sphere  made  by  a  plane  through  the 
center  of  the  sphere  is  called  a  great  circle. 

A  section  of  a  sphere  made  by  a  plane  that  does  not 
pass  through  the  center  of  the  sphere  is  called  a  small  circle. 


82  SOLID   GEOMETRY 

The  diameter  of  a  sphere  perpendicular  to  the  plane  of 
a  circle  of  a  sphere  is  called  the  axis  of  that  circle. 

The  ends  of  axis  of  a  circle  of  a  sphere  are  called  the  poles 
of  that  circle. 

THEOREMS   CONCERNING   CIRCLES   ON   SPHERES 

102.  THEOREM  64.  The  axis  of  a  circle  of  a  sphere 
passes  through  the  center  of  that  circle. 

Suggestion.  Th.  64  is  a  corollary  of  Th.  63.  In  Th.  63,  OX  was 
constructed  from  O_L  the  plane  of  the  circle  and  meeting  the  plane  at 
X.  The  point  X  was  proved  to  be  the  center  of  the  circle.  Why  is 
OX  the  axis  of  the  circle  ?  Use  Fig.  103,  No.  1. 

THEOREM  65.  All  great  circles  on  a  given  sphere  are 
congruent. 

THEOREM  66.  Any  two  great  circles  on  a  sphere  bisect 
each  other. 

Suggestion.  Show  that  the  intersection  of  the  planes  of  the  two 
great  circles  is  a  diameter  of  each  circle. 

THEOREM  67.  Every  great  circle  on  a  sphere  bisects  the 
sphere. 

THEOREM  68.  Three  points  on  a  sphere  no  two  of  which 
are  at  the  ends  of  a  diameter  determine  a  small  circle  of 
a  sphere. 

THEOREM  69.  Two  points  on  a  sphere  not  the  ends  of 
a  diameter  determine  a  great  circle  of  a  sphere. 

The  length  of  the  minor  arc  of  a  great  circle  on  a  sphere 
joining  two  given  points  on  the  sphere  is  called  the  spherical 
distance  between  the  two  given  points.  This  is  the  shortest 
distance  between  the  two  points  on  the  spherical  surface. 

Ex.  1.  How  many  great  circles  of  a  sphere  can  pass  through 
the  opposite  ends  of  a  diameter?  Why? 

Ex.  2.  How  many  small  circles  of  a  sphere  can  pass  through 
two  points  on  a  sphere  if  the  two  points  are  (1)  the  opposite  ends 
of  a  diameter;  (2)  not  the  opposite  ends  of  a  diameter? 


THE  SPHERE 


83 


POLAR  DISTANCE 

103.  THEOREM  70.    Every  point  on  a  circle  on  a  sphere 
is  equally  distant  from  each  of  its  poles. 


,Vo.  1 


No.  2 


FIG.  103 


Hypothesis:  0  is  any  sphere.  C  is  a  circle  on  the  sphere. 
P  and  P'  are  the  poles  of  circle  C.  A  and  B  are  any  two 
points  on  circle  C. 

Conclusion:  A  and  B  are  equally  distant  from  P  and  also 
from  P'. 

Analysis  and  construction: 

A.  I.  To  prove  A  and  B  equally  distant  from  P,  draw 

minor  arcs  of  great  circles  joining  AP  and  BP 

X*^N  /"""^S 

and  prove  AP  =  BP. 
II.    /.  draw  PA  and  PB  and  prove  AP  =  BP. 

B.  I.  To  prove  A  and  B  equally  distant  from  P',  draw 

minor  arcs  of  great  circles  joining  AP1  and 
BP'  and  prove  AP'  =  BP'. 

The  spherical  distance  from  the  nearer  pole  of  a  circle 
of  a  sphere  to  any  point  on  the  circle  is  called  the  polar 
distance  of  the  circle. 

COR.  The  polar  distance  of  a  great  circle  of  a  sphere 
is  a  quadrant,  or  an  arc  of  90°. 

NOTE.  The  proof  of  Th.  70  shows  that  the  theorem  is  true  if  dis- 
tance means  either  (I)  rectilinear  distance  or  (2)  spherical  distance. 


84 


SOLID  GEOMETRY 


NOTE.  In  order  that  a  true  conception  of  much  of  the  work  in 
this  chapter  may  be  obtained  a  slated  globe  should  be  used  in  class, 
and  the  pupil  should  prepare  his  lessons  with  a  ball  on  which  lines 
may  be  drawn.  To  draw  a  circle  on  a  sphere,  it  is  evident  that  its 
pole  and  the  polar  distance  or  the  linear  distance  must  be  known. 

104.  THEOREM  71.  If  a  point  is  a  quadrant's  distance 
from  each  of  two  points  in  a  sphere,  it  is  the  pole  of  the  great 
circle  passing  through  these  two  points. 


Hypothesis:  0  is  any  sphere.  PA  and  PB  are  quadrants. 
.  Conclusion:  P  is  the  pole  of  the  great  circle  ABC  through 
points  A  and  B. 

Analysis  and  construction: 
I.  To  prove  P  the  pole  of  great  circle  ABC,  join  PO  and 

prove  PO  _L  plane  ABC. 
II.  To  prove  PO  J_  plane  ABC,  join  OA  and  BO  and 

prove  /.POA  and  /.POB  right  angles. 

MISCELLANEOUS   EXERCISES 

105.  1.  If  a  plane  is  tangent  to  a  sphere,  the  plane  of  every 
great  circle  through  the  point  of  contact  is  perpendicular  to  the 
tangent  plane. 

2.  A  line  perpendicular  to  the  plane  of  a  circle  of  a  sphere  at 
the  center  of  the  circle  passes  through  the  center  of  the  sphere. 

3.  A  segment  joining  the  center  of  a  circle  on  a  sphere  with  the 
center  of  the  sphere  is  perpendicular  to  the  plane  of  the  circle. 


THE  SPHERE  85 

4.  If  the  planes  of  two  great  circles  on  a  sphere  are  perpen- 
dicular to  each  other,  each  circle  passes  through  the  pole  of  the 
other. 

5.  Points  A  and  B  are  at  the  opposite  ends  of  a  diameter  of  a 
sphere,  and  a  given  point  P  is  at  a  quadrant's  distance  from  both 
.4  and  B.     Of  which  great  circle  or  circles  through  A  and  B  is  P 
the  pole? 

6.  Into  how  many  parts  do  three  great  circles  of  a  sphere 
divide  the  surface  of  the  sphere?     Discuss  two  cases. 

7.  The  sections  of  a  sphere  made  by  parallel  planes  have  the 
same  poles. 

8.  State  and  prove  the  converse  to  Ex.  7. 

9.  If  two  planes  cut  a  sphere  at  equal  distances  from  the  center 
of  the  sphere,  the  circles  thus  formed  are  congruent. 

Analysis  and  construction: 
I.  To  prove  the  circles  congruent,  prove  their  diameters  equal. 

II.  Let  OA  and  OB  be  the  perpendiculars  from  the  center  of  the 
sphere  to  the  planes  of  the  two  circles.  The  plane  of  OA 
and  OB  will  cut  the  planes  of  the  two  circles  in  diameters. 
Use  plane  geometry. 

10.  State  and  prove  the  converse  to  Ex.  9. 

11.  On  the  same  sphere  or  on  congruent  spheres  congruent 
circles  have  equal  polar  distances. 

12.  State  and  prove  the  converse  to  Ex.  11. 

13.  The  plane  of  a  circle  of  a  sphere  is  perpendicular  to  the 
planes  of  all  great  circles  passing  through  its  poles. 

14.  A  plane  that  bisects  at  right  angles  a  chord  of  a  sphere 
contains  the  center  of  the  sphere. 

15.  Show  that  a  sphere  can  be  inscribed  in  a  given  cube.     Show 
that  one  can  be  circumscribed  about  a  given  cube. 

16.  Given  a  segment  equal  to  the  edge  of  a  cube.     Construct 
on  a  plane  a  segment  equal  to  the  radius  of  the  inscribed  sphere, 
and  one  equal  to  the  radius  of  the  circumscribed  sphere. 

17.  It  can  be  proved  that  the  six  planes  bisecting  the  dihedral 
angles  of  any  tetrahedron  meet  in  a  point.     Show  from  this  how 
to  inscribe  a  sphere  in  any  given  tetrahedron. 


86 


SOLID  GEOMETRY 


SPHERICAL  ANGLES 


106.  If  two  great  circle  arcs  on  a  sphere  intersect,  they 
are  said  to  form  a  spherical  angle.     The  point  of  intersection 
is  called  the  vertex  of  the  angle.     The  two  great  circle  arcs 
are  called  the  sides  of  the  angle. 

The  measure  of  a  spherical  angle  is  defined  as  the  measure 
of  the  plane  angle  made  by  the  tangents  to  the  two  arcs  at 
their  intersection. 

107.  THEOREM   72.     A   spherical   angle   has   the    same 
measure  as  (1)  the  dihedral  angle  formed  by  the  planes  of 
its  sides,  and  (2)  the  arc  of  a  great  circle  drawn  with  its 
vertex  as  a  pole  and  included  between  its  sides. 


FIG.  105 

Hypothesis:  AXB  is  a  spherical  angle  formed  by  the 
two  great  circle  arcs  AX  and  BX  on  sphere  O.  The  planes 
oi  AX  and  BX  intersect,  forming  the  dihedral  angle  whose 

edge  is  XO.     AB  is  the  arc  of  a  great  circle  drawn  with  X 
as  pole  and  included  between  the  sides  of  Z  AXB.     XY  and 

XZ  are  tangents  to  AX  and  BX  at  X. 
Conclusion: 

a.  /.AXB  has  the  same  measure  as  dihedral  angle  XO. 

b.  /  AXB  has  the  same  measure  as  AB. 

Analysis  for  a:  To  prove  that  /.AXB  has  the  same 
measure  as  dihedral  angle  XO,  prove  that  /.  YXZ  is  the 
plane  angle  of  the  dihedral  angle  XO. 


THE   SPHERE  87 

Analysis  and  construction  for  b: 

I.  To  prove  that  Z.AXB  has  the  same  measure  as  AB, 
join  AO  and  BO  and  prove  Z  YXZ=  Z.AOB,  and 
Z.AOB  has  the  same  measure  as  AB. 

II.  To  prove    4YXZ=  Z.AOB,  prove  XY\\OA,  and 
XZ    OB. 


III.    /.  prove  OA  and  OJS  _L  XO  at  0. 

COR.  I.  Two  great  circle  arcs  are  perpendicular  to  each 
other  if  their  planes  are  perpendicular  to  each  other. 

COR.  II.  If  two  great  circle  arcs  are  perpendicular  to 
each  other,  then  each  passes  through  the  pole  of  the  other. 

Ex.  1.  If  an  arc  of  one  great  circle  passes  through  the  pole  of 
the  second,  the  two  arcs  are  perpendicular  to  each  other. 

Ex.  2.  If  one  great  circle  arc  passes  through  the  pole  of  a  second, 
then  the  second  passes  through  the  pole  of  the  first. 

Ex.  3.  If  two  points  on  a  sphere  are  a  quadrant's  distance 
apart,  then  each  is  the  pole  of  one  great  circle  passing  through  the 
other. 

SPHERICAL  TRIANGLES 
LINES   ON   SPHERES 

108.  We  have  stated  that  the  shortest  distance  between 
two  points  on  the  surface  of  a  sphere  is  the  minor  arc  of  a 
great  circle  passing  through  these  points.  This  fact  is 
important  for  two  reasons. 

I.  It  is  important  theoretically.  Arcs  of  great  circles 
on  a  sphere  take  the  place  in  our  work  of  straight  lines  on  a 
plane.  As  we  study  figures  made  up  of  straight  lines  on  a 
plane  in  plane  geometry,  so  we  are  to  study  figures  made  up 
of  arcs  of  great  circles  on  a  sphere  in  solid  geometry.  The 
geometry  of  these  figures  can  be  studied  by  the  use  of  the 
same  methods  as  those  used  in  studying  triangles  and  poly- 
gons on  a  plane,  but  this  method  is  out  of  place  here. 


88  SOLID   GEOMETRY 

II.  It  is  important  practically.  Use  is  made  of  it  in 
sea  sailing  and  airplane  work.  To  obtain  the  answers  to 
the  following  questions,  use  a  small  geographical  globe. 
Suppose  the  earth  to  be  a  perfect  sphere. 

Ex.  1.  What  kinds  of  circles  are  the  parallels  of  latitude  on 
the  earth?  What  kinds  of  circles  are  the  meridians?  What  kind 
of  a  circle  is  the  equator. 

Ex.  2.  Of  what  circle  is  the  North  Pole  of  the  earth  the  pole? 

Ex.  3.  What  is  the  shortest  route  from  the  Pacific  end  of  the 
Panama  Canal  to  Japan?  Stretch  a  string  between  the  two  points 
on  the  globe. 

Ex.  4.  What  is  the  shortest  route  from  New  York  to  England? 
From  Cape  Town,  South  Africa,  to  Melbourne,  Australia? 

Ex.  5.  What  is  meant  by  the  statement  that  there  are  no 
lines  on  a  sphere  that  correspond  to  parallel  straight  lines  on  a 
plane? 

Ex.  6.  Why  are  the  parallels  of  latitude  on  the  earth  called 
parallels? 

DEFINITIONS 

109.  A  closed  figure  formed  of  two  or  more  arcs  of  great 
circles  on  a  sphere,  no  one  of  which  is  greater  than  a  semi- 
great  circle,  is  called  a  spherical  polygon. 

NOTE.  Spherical  polygons  can  be  constructed  in  which  one  or 
more  sides  are  greater  than  a  semi-great  circle.  Such  polygons  are 
called  general  spherical  polygons  and  will  not  be  considered  in  this 
book. 

The  intersection  of  the  arcs  are 
called  the  vertices  of  the  polygon. 
In  Fig.  106,  A,  B,C,  and  D  are  the 
vertices  of  the  spherical  polygon 
ABCD. 

The  arcs  are  called  the  sides  of 
the  spherical  polygon.  In  Fig.  106,  FIG.  106 

AB,  BC,  CD,  and  DA  are  the  sides  of  the  polygon  ABCD. 


THE   SPHERE 


FIG.  107 


The  spherical  angles  formed  by  the  arcs  are  called  the 
angles  of  the  spherical  polygon.  What  are  the  angles  of 
the  spherical  polygon  ABCD  in  Fig.  106? 

A  spherical  polygon  of  two  sides 
is  called  a  lune.  In  this  case  the 
sides  of  the  spherical  polygon  are 
semi-great  circles.  'In  Fig.  107,  the 

figure  formed  by  ACB  and  ADB  is  a 
lune.  There  is,  obviously,  no  figure 
on  a  plane  that  corresponds  to  a  lune 
on  the  sphere. 

A  spherical  polygon  of  three  sides " 
is    called    a    spherical    triangle.     In 
Fig.  108,  ABC  is  a  spherical  triangle. 

Spherical  triangles  are  isosceles  or 
equilateral  as  in  plane  geometry. 

A  convex  spherical  polygon  is  one 
in  which  if  any  side  is  extended  the 
whole  polygon  lies  on  one  side  of  the 
extended  arc.     Unless   otherwise   stated,  convex  spherical 
polygons  are  intended. 

Exercise.  Using  a  slated  globe  or  a  ball  of  some  kind  on  which 
marks  can  be  made,  draw  a  spherical  triangle  with  one  side  greater 
than  a  semi-great  circle.  Is  it  concave  or  convex?  Draw  one  in 
which  two  sides  are  greater  than  semi-great  circles.  Can  you 
draw  one  in  which  two  sides  are  semi-great  circles?  Why? 

SPHERICAL   POLYGONS   AND    CENTRAL 
POLYHEDRAL   ANGLES 

110.  A  polyhedral  angle  with  its  vertex  at  the  center  of 
the  sphere  cuts  the  surface  of  the  sphere  in  a  convex  spherical 
polygon.  Conversely,  if  the  vertices  of  a  convex  spherical 
polygon  are  joined  to  the  center  of  the  sphere,  a  polyhedral 
angle  is  formed  with  its  vertex  at  the  center  of  the  sphere. 
This  polyhedral  angle  is  called  the  corresponding  central 
polyhedral  angle  of  the  spherical  polygon. 
7  • 


FIG.  108 


90  SOLID   GEOMETRY 

NOTE.  The  theorems  that  we  have  studied  about  polyhedral 
angles  are  for  convex  polyhedral  angles  only.  This  is  one  reason  why 
we  are  confining  our  study  of  spherical  polygons  to  those  whose  sides 
are  less  than  180°  and  to  convex  figures. 

To  each  side  of  the  spherical  polygon  corresponds  one 
face  angle  of  the  polyhedral  angle,  namely:  that  face  angle 
that  is  subtended  by  the  side  of  the  polygon.  To  each 
angle  of  the  spherical  polygon  corresponds  one  dihedral  angle 
of  the  polyhedral  angle,  namely :  that  dihedral  angle  which 
is  formed  by  the  planes  of  the  sides  of  the  spherical  angle. 
The  next  two  theorems  follow  at  once. 

THEOREM  73.  Any  side  of  a  spherical  polygon  has  the 
same  measure  as  the  corresponding  face  angle  of  the  corre- 
sponding central  polyhedral  angle.  (See  Fig.  109.) 

Suggestion.  As  the  sides  of  a  spherical  polygon  are  arcs,  they  may 
be  measured  in  degrees. 

THEOREM  74.  Any  angle  of  a  spherical  polygon  has  the 
same  measure  as  the  corresponding  dihedral  angle  of  the 
corresponding  central  polyhedral  angle.  (See  Fig.  109.) 

Suggestion.     This  is  a  restatement  of  what  theorem? 

SOME  PROPERTIES   OF   SPHERICAL   TRIANGLES 

111.  THEOREM  75.  The  sum  of  two  sides  of  a  spherical 
triangle  is  greater  than  the  third  side. 


FIG.  109 

Suggestion.     The  proof  follows  at  once  from  Ths.  73  and  30 


THE  SPHERE  91 

Some  theorems  concerning  spherical  polygons  bear  a 
certain  peculiar  relation  to  the  corresponding  theorems 
concerning  polyhedral  angles.  This  relation  is  called 
reciprocal  and  follows  at  once  from  Ths.  73  and  74.  When 
the  measures  of  arcs  and  angles  are  concerned,  one  theorem 
may  be  obtained  from  the  other  by  the  substitution  of  the 
measure  of  the  sides  of  the  polygon  and  the  measure  of  the 
corresponding  face  angles  of  the  polyhedral  angle  for  each 
other,  or  the  measure  of  the  spherical  angles  and  the  meas- 
ure of  the  corresponding  dihedral  angles  for  each  other. 

THEOREM  76.  The  sum  of  the  sides  of  a  spherical 
polygon  is  less  than  a  great  circle. 

Ex.  1.  Describe  the  corresponding  central  trihedral  angle  of  a 
spherical  triangle  that  has  (1)  two  sides  equal;  (2)  three  sides 
equal;  (3)  two  angles  equal;  (4)  three  angles  equal.  Give  reasons. 

Ex.  2.  Describe  the  spherical  triangle  that  would  correspond 
to  a  central  trihedral  angle  that  has  three  right  dihedral  angles. 
Can  you  draw  this  triangle  on  the  surface  of  a  given  sphere  such 
as  a  slated  globe? 

Ex.  3.  Can  you  answer  Ex.  2  if  the  initial  trihedral  angle  has 
only  two  right  dihedral  angles?  Two  obtuse  dihedral  angles? 

POLAR   TRIANGLES 

112.  If  A,  B,  and  C  are  the  vertices  of  a  spherical  tri- 
angle, and  a,  b,  and  c  are,  respectively,  the  sides  opposite 
these  vertices,  and  if  A'  is  that  pole 
of  side  a  that  is  on  the  same  side  of  a 
as  the  vertex  A,  B'  of  b,  and  C'  of  c, 
then  A'B'C'  is  called  the  polar  triangle 
of  AABC.  In  Fig.  110,  A'  and 
A"  are  the  poles  of  the  side  a  in 
AA5C.  A  and  A'  are  on  the  same 
side  of  a.  Then  A'  is  one  vertex  of  FIG.  110 

the  polar  triangle  of  AABC.     In  the  same  way  the  other 
vertices  are  C'  and  B'. 


92 


SOLID   GEOMETRY 


113.  THEOREM  77.  If  one  spherical  triangle  is  the  polar 
of  a  second,  then  reciprocally  the  second  is  the  polar  of 
the  first.  » 


Q 


FIG.  Ill 

Hypothesis:  ABC  is  a  spherical  triangle.  A'B'C'  is  UK- 
polar  triangle  of  A  ABC. 

Conclusion:     ABC  is  the  polar  triangle  of  A  A'B'C'. 
Analysis: 

I.  To  prove  ABC  the  polar  of  A  A'B'C,  prove  A  the 
pole  of  B'C,  B  the  pole  of  A'C',  and  C  the  pole  of 
B'A'. 
II.  To  prove  A  the  pole  of  B'C ',  prove  A  a  quadrant's 

distance  from  B'  and  C ' . 
Outline  of  proof: 

B'  is  the  pole  of  AC. 
.'.  B'A  is  a  quadrant. 
C'  is  the  pole  of  AB. 
.'.  C'A  is  a  quadrant. 
/.  A  is  the  pole  of  B'C: 
Similarly 

B  is  the  pole  of  A'C'. 
C  is  the  pole  of  B'A'. 
:.  ABC  is  the  polar  of  A'B'C'. 

Ex.  1.  Show  how  to  construct  on  a  ball  or  spherical  black- 
board two  polar  spherical  triangles. 

Ex.  2.  Construct  on  a  ball  or  spherical  blackboard  the  polar 
of  a  spherical  triangle  that  has  each  side  less  than  a  quadrant. 


THE   SPHERE 


93 


Ex.  3.  Construct  as  in  EX.  2  the  polar  of  a  triangle  that  has 

a.  One  side  greater  and  two  sides  less  than  a  quadrant. 

b.  Two  sides  greater. and  one  side  less  than  a  quadrant. 

c.  Each  side  greater  than  a  quadrant. 

d.  One  side  a  quadrant. 

e.  Two  sides  quadra'nts. 
/.   Three  sides  quadrants. 

Ex.  4.  Describe  the  corresponding  central  trihedral  angle  of 
each  triangle  mentioned  in  Ex.  3  and  of  each  polar. 

114.  THEOREM  78.  In  two  polar  spherical  triangles  the 
sum  of  the  measures  of  any  angle  of  one  and  that  side  of 
the  other  of  which  it  is  a  pole  is  180°. 


C 


FIG.  112 

Hypothesis:     ABC  and  A'B'C'  are  two  polar  spherical 
triangles.     Let  A  represent  the  measure  of  ZA,  and  a'  the 

measure  of  B'C'. 

Conclusion:     A  +af  =  180°. 
Analysis  and  construction: 

I.  To  prove  A  -fa'  =  180°,  compare  A  with  an  arc  which 
is  the  supplement  of  a1 '. 

II.    /.  continue  AB  and  AC  to  meet  C'B'  in  X  and  Y 
respectively.     Let  m  represent  the  measure  of  XV. 
Prove  (1)          'A=m. 

(2)    m+a'  =  180°. 

III,  To  prove  ra+a'  =  180°,  prove 

(1)  C'X+B'Y  =  C'B'+XY 

(2)  C'X+B'Y=m°. 

IV.  /.  prove  C'X  =  90°,  and  #'F  =  90°. 


94  SOLID   GEOMETRY 

COR.  I.  On  the  same  sphere  or  on  congruent  spheres, 
if  two  spherical  triangles  are  mutually  equilateral,  their 
polars  are  mutually  equiangular  (Fig.  113). 

COR.  II.  On  the  same  sphere  or  on  congruent  spheres, 
If  two  spherical  triangles  are  mutually  equiangular,  their 
polars  are  mutually  equilateral  (Fig.  113). 


FIG.  113 

Exercise.  The  sides  of  a  spherical  triangle  are  65°,  115°,  and 
120°.  Find  the  number  of  degrees  in  each  angle  (1)  of  its  polar, 
(2)  of  each  of  the  triangles  formed  by  the  intersection  of  the  great 
circles  of  which  the  sides  of  the  polar  are  arcs. 

THE   SUM   OF  THE   ANGLES   OF  A   SPHERICAL  TRIANGLE 

116.  THEOREM  79.  The  sum  of  the  angles  of  a  spherical 
triangle  is  more  than  two  and  less  than  six  right  angles. 


FIG.  114 

Hypothesis:     A  ABC  is  any  spherical  triangle.     A,   B, 
and  C  represent  the  measures  of  AA,B,  and  C  respectively. 

Conclusion:     A+B+C  >  2  rt.  A  and  <  6  rt.  A 


THE   SPHERE  95 

Analysis  and  construction:  To  find  the  value  of  A  +B+ 
C,  construct  the  polar  of  &ABC.  Let  a',  bf,  and  cr  represent 
the  measures  of  the  sides  .of  the  polar.  Compare  A  +  B+C 
with  a'+b'+c'. 

Outline  of  proof: 


=  180°. 


/.    A+B+C+a'+b'+c'  =  5ttf    or    6    rt.     A, 
but  a'+b'+c'  >Q. 

.'.  A+B+C<6rt.  A. 
II.  Again,  a'+b'+c'  <  4  rt.  A  . 


In  giving  the  reasons  note  the  application  of  the  inequality  assump- 
tions. 

Ex.  1.  On  a  ball  or  a  spherical  blackboard  draw  a  spherical 
triangle  with  two  right  angles;  with  three  right  angles;  with  two 
obtuse  angles. 

A  spherical  triangle  with  two  right  angles  is  called  a 
birectangular  spherical  triangle.  One  with  three  right  angles 
is  called  a  trirectangular  spherical  triangle. 

The  number  of  degrees  by  which  the  sum  of  the  angles 
of  a  spherical  triangle  exceeds  180°  is  called  the  spherical 
excess  of  the  triangle. 

Ex.  2.  What  can  you  say  concerning  the  sum  of  the  measures 
of  the  dihedral  angles  of  a  trihedral  angle?  Why?  What  do  you 
know  concerning  the  sum  of  the  face  angles  of  a  trihedral  angle? 

Ex.  3.  In  a  birectangular  spherical  triangle  the  sides  opposite 
the  right  angles  are  quadrants. 

Ex.  4.  The  sides  of  a  trirectangular  spherical  triangle  are 
quadrants. 

Ex.  5.  The  polar  triangle  of  a  birectangular  spherical  triangle 
is  birectangular. 

Ex.  6.  What  is  the  polar  of  a  trirectangular  spherical  triangle? 
Why? 


96  SOLID   GEOMETRY 

CONGRUENT   AND    SYMMETRIC   SPHERICAL   TRIANGLES 

116.  Two  spherical  triangles  on  the  same  sphere  or  on 
congruent  spheres  are  said  to  be  congruent  if  the  sides  and 
angles  of  one  are  equal  respectively  to  the  sides  and  angles 
of  the  other  and  arranged  in  the  same  order. 

As.  24.  Two  spherical  triangles  or  two  trihedral  angles 
congruent  to  a  third  are  congruent  to  each  other. 

The  next  two  theorems  follow  at  once  from  the  defini- 
tion above.  See  the  remark  on  p.  91  concerning  the  proof 
of  Th.  75. 

What  are  congruent  trihedral  angles?  (§52.) 
THEOREM  80.     On  the  same  sphere  or  on  congruent 
spheres,  if  two  spherical  triangles  are  congruent,  the  corre- 
sponding central  trihedral  angles  are  congruent  (Fig.  115). 


FIG.  115  FIG.  116 

THEOREM  81.  In  the  same  sphere  or  in  congruent 
spheres,  if  two  central  trihedral  angles  are  congruent,  the 
corresponding  spherical  triangles  are  congruent  (Fig.  115). 

THEOREM  82.  On  the  same  sphere  or  on  congruent 
spheres,  if  two  spherical  triangles  are  congruent,  their 
polar  triangles  are  congruent  (Fig.  116).  (See  Th.  78.) 

117.  Two  spherical  triangles  on  the  same  sphere  or  on 
congruent  spheres  are  said  to  be  symmetric  if  the  sides  and 
angles  of  one  are  equal  respectively  to  the  sides  and  angles 
of  the  other  and  arranged  in  the  opposite  order. 
What  are  symmetric  trihedral  angles?     (§52.) 
In  general,  two  symmetrical  spherical  triangles  or  two 
symmetric  trihedral  angles  cannot  be  made  to  coincide. 


THE   SPHERE 


97 


Ex.  1  Show  how  two  plane  figures  that  are  symmetric  with 
respect  to  a  point  or  to  a  line  can  be  made  to  coincide. 

Ex.  2.  Draw  on  a  spherical  blackboard  figures  to  illustrate 
two  congruent  and  two  symmetric  spherical  triangles. 

Ex.  3.  Construct  from  cardboard  two  symmetric  trihedral 
angles,  making  the  faces  circular  sectors  with  equal  radii. 

As.  .25.  Two  spherical  triangles  or  two  trihedral  angles 
symmetric  to  a  third  are  congruent  to  each  other. 

The  next  two  theorems  follow  at  once  from  the  above 
definition  and  the  remarks  following.  Th.  75. 

THEOREM  83.  On  the  same  sphere  or  on  congruent 
spheres,  if  two  spherical  triangles  are  symmetric,  the 
corresponding  central  trihedral  angles  are  symmetric 
(Fig.  117). 


FIG.  117 

THEOREM  84.  In  the  same  sphere  or  in  congruent 
spheres,  if  two  central  trihedral  angles  are  symmetric,  the 
corresponding  spherical  triangles  are  symmetric  (Fig.  117). 

THEOREM  85.  On  the  same  sphere  or  on  congruent 
spheres,  if  two  spherical  triangles  are  symmetric,  then- 
polar  triangles  are  symmetric  (Fig.  118). 

118.  If  the  edges  of  one  trihedral  angle  are  prolongations 
of  the  edges  of  another  trihedral  angle,  the  trihedral  angles 
are  said  to  be  vertical  trihedral  angles,  and  the  correspond- 
ing spherical  triangles  cut  out  on  the  sphere  are  said  to  be 
vertical  spherical  triangles. 

THEOREM  86.  Two  vertical  trihedral  angles  are  sym- 
metric. 


98 


SOLID  GEOMETRY 


THEOREM  87. 
metric  (Fig.  119). 


Two  vertical  spherical  triangles  are  sym- 

B 


FIG.  119 

119.  THEOREM  88.  If  two  spherical  triangles  on  the 
same  or  on  congruent  spheres  have  two  sides  and  the  in- 
cluded angle  of  one  equal  to  two  sides  and  the  included 
angle  of  the  other, 

a.  The  triangles  are  congruent  if  the  parts  are  arranged 
in  the  same  order. 

b.  The  triangles  are  symmetric  if  the  parts  are  arranged 
in  the  opposite  order. 


FIG.  120 

Hypothesis:    ABC  and  A'B'C'  are  two  spherical  triangles 
in  which  Z A  =  Z.A',  b  =  b',  and\  =  c'. 
Conclusion: 

a.  If  /.A,  b,  and  c  are  arranged  in  the  same  order  as 

ZA',  b',  and  c',   AABC^AA'B'C'  (AI^AII). 

b.  If  /.A,b,  and  c  are  arranged  in  the  opposite  order  to 

ZA',  b1,  and  c',  AABC  and  AA'B'C'  are  sym- 
metric (All  symmetric  to  Alii). 

Analysis  for  a:    To  prove  A I  ^  AH,  show  that  they  will 
coincide  if  superposed  as  in  plane  geometry. 


THE   SPHERE  99 

Analysis  and  construction  for  b: 
I.  To    prove    AIII   symmetric   to    All,   prove    AIII 

congruent  to  a  triangle  that  is  symmetric  to  All. 
II.    .*.  construct  A IV  symmetric  to  All  and  prove  A IV 

congruent  to  AIII. 
NOTE.     AIV  is  not  shown  in  the  figure. 

COR.  If  two  trihedral  angles  have  two  face  angles  and 
the  included  dihedral  angle  of  one  equal  to  two  face  angles 
and  the  included  dihedral  angle  of  the  other,  the  trihedral 
angles  are  congruent  if  the  parts  are  arranged  in  the  same 
order,  and  symmetric  if  they  are  arranged  in  the  opposite 
order. 

120.  THEOREM  89.  If  two  spherical  triangles  on  the 
same  or  on  congruent  spheres  have  two  angles  and  the 
included  side  of  one  equal  to  two  angles  and  the  included 
side  of  the  other, 

a.  The  triangles  are  congruent  if  the  parts  are  arranged 
hi  the  same  order. 

b.  The  triangles  are  symmetric  if  the  parts  are  arranged 
in  the  opposite  order. 

Analysis  for  a: 

I.  To  prove  A I  ^  All,  prove  that  their  polars  are  con- 
gruent. 

II.  To  prove  the  polars  congruent,  prove  that  they  have 
two  sides  and  the  included  angle  of  one  equal 
to  A,  etc. 

Analysis  for  b:  To  prove  AIII  symmetric  to  All,  prove 
that  their  polars  are  symmetric. 

COR.  If  two  trihedral  angles  have  two  dihedral  angles 
and  the  included  face  angle  of  one  equal  to  two  dihedral 
angles  and  the  included  face  angle  of  the  other,  the  trihedral 
angles  are  congruent  if  the  parts  are  arranged  in  the  same 
order,  and  symmetric  if  they  are  arranged  in  the  opposite 
order. 


100  SOLID  GEOMETRY 

121.  THEOREM  90.  If  two  spherical  triangles  on  the 
same  or  on  congruent  spheres  have  three  sides  of  one  equal 
to  three  sides  of  the  other,  the  triangles  are  either  con- 
gruent or  symmetric. 

C'  ,  C  .  C 


FIG.  121 

Analysis:  To  prove  &ABC  congruent  or  symmetric 
to  AA'B'C',  prove  that  the  corresponding  central  trihedral 
angles  are  congruent  or  symmetric.  (See  Th.  38,  Cor.) 

122.  THEOREM  91.     If  two  spherical  triangles  on  the  same 
or  on  congruent  spheres  have  three  angles  of  one  equal 
to  the  three  angles  of  the  other,  the  triangles  are  either 
congruent  or  symmetric. 

Analysis:  To  prove  A  ABC  congruent  or  symmetric  to 
AA'B'C',  prove  their  polars  congruent  or  symmetric. 

COR.  If  two  trihedral  angles  have  the  three  dihedral 
angles  of  one  equal  to  the  three  dihedral  angles  of  the  other, 
the  trihedral  angles  are  either  congruent  or  symmetric. 

ISOSCELES   SPHERICAL   TRIANGLES 

123.  THEOREM  92.    The  angles  opposite  the  equal  sides 
of  an  isosceles  spherical  triangle  are  equal. 


Suggestion.  Join  the  vertex  C  with  X, 
the  mid-point  of  the  base  AB,  by  an  arc  of 
a  great  circle.  Prove  Z.A  and  Z.B  corre- 
sponding angles  of  symmetric  spherical  tri- 
angles (Fig.  122). 


FIG.  122 


THE   SPHERF. 


THEOREM  93.  If  two  angles  of  a  spherical  triangle  are 
equal,  the  sides  opposite  the  equal  angles  are  equal  (Fig. 
122). 

Analysis:     To  prove  AC—  CB,  prove  that  A  A'  and  B'  of 
the  polar  are  equal. 

Ex.  1.  If  one  of  two  polar  spherical  triangles  is  isosceles,  the 
other  is. 

Ex.  2.  If  a  spherical  triangle  is  equilateral,  its  polar  is  also 
equilateral. 

THEOREM  94.  If  two  symmetric  spherical  triangles  are 
isosceles,  they  are  congruent. 


FIG.  123 

Hypothesis:  ABC  and  A'B'C'  are  two  symmetric 
isosceles  spherical  triangles  with  AB  =  A'B',  AC=A'C', 
Cll  =  C'B',  Z6 

Conclusion : 

Analysis:     To    prove    AABC  ^  AA'B'C',    prove   AC 
C'B',    ZC=  ZC',  CB  =  C'A'. 


EQUIVALENT    SPHERICAL   TRIANGLES 

124.  We  have  seen  that,  in  general,  two  symmetric 
spherical  triangles  cannot  be  made  to  coincide.  Why? 
We  will  prove,  however,  that  two  symmetric  spherical  tri- 
angles are  equivalent;  that  is,  they  cover  the  same  extent 
of  spherical  surface.  In  general,  we  have 

As.  26.  Two  spherical  polygons  are  equivalent  if  they 
are  congruent  or  are  made  up  of  parts  congruent  in  pairs. 


102  bCH  ,ID  GEOMETRY 

THEOREM  95.     Two  symmetric  spherical  triangles  are 
equivalent. 


FIG.  124 

Hypothesis:     AABC  and  AA'B'C'  are  two  symmetric 
spherical  triangles. 

Conclusion:     AABC=  AA'B'C'. 
Analysis  and  construction: 
I.  To  prove  AABC  =  AA'B'C',  prove  that  they  can 

be  divided  into  parts  congruent  in  pairs. 
II.    .'.    divide     &ABC    and    AA'B'C'    into    isosceles 
spherical  triangles  symmetric  in  pairs. 

III.  ,*.    find  X  and  X1  ',  the  poles  of  the  small  circles 

through  A,  B,  and  C  and  through  A',  B',  and  C' 

respectively.     Draw  AX,  BX,  CX,  A*X',  &X', 

CX',  and  prove  AXAB  and  AX'A'B',  AXBC 
and  AX'£'C',  also  &XAC  and  X'A'C't  isosceles 
and  symmetric. 

IV.  /.  prove  AX  =BX  =  CX  =  A^Xf  =  B^Xf  =  CX'  . 
V.    .'.  prove  QABC&QA'B'C. 

VI.    .'.  prove  plane  AABC  g  plane 


SUPPLEMENTARY  EXERCISES 

MISCELLANEOUS  EXERCISES 

125.  1.  If  two  sides  of  a  spherical  triangle  are  quadrants,  the 
included  angle  has  the  same  measure  as  the  third  side. 

2.  The  spherical  excess  of  a  birectangular  spherical  triangle  is 
the  measure  of  the  third  angle. 


THE   SPHERE 


103 


3.  Find  the  locus  of  poles  of  small  circles  that  pass  through 
two  given  points  in  a  sphere. 

4.  Find  the  locus  of  poles  of  great  circles  that  pass  through 
the  ends  of  a  diameter  on  a  sphere. 

5.  The  central  trihedral  angles  that  correspond  to  two  polar 
spherical  triangles  have  the  edges  of  one  perpendicular  to  the 
faces  of  the  other. 

6.  Find  the  diameter  of  a  given  material  sphere. 


Nd.  1  No.  2 

FIG.  125 
A  nalysis: 

I.  We  can  find  the  diameter  of  the  sphere  if  we  can  construct  on 
paper  or  blackboard  a  circle  equal  to  a  great  circle  of  the 
sphere  and  find  its  diameter. 

II.  To  construct  a  circle  equal  to  a  great  circle  of  the  sphere, 

locate  three  points  on  this  circle.     These  three  points  may 

be  the  ends  of  a  diameter  of  any  small  circle  and  its  pole 

(Fig.  125,  No.  3). 

III.    .'.  construct  a  circle  equal  to  any  small  circle  on  the  sphere. 

Construction: 

I.  .'.  with  any  point,  P,  as  a  pole  draw  any  small  circle  on  the 
sphere.  Draw,  on  paper  or  blackboard,  AABC  (Fig.  125, 
No.  2)  whose  sides  are  equal  to  the  chords  AB,  BC,  CA 
(No.  1 ) .  (The  segments  A  B,  B  C,  and  CA  may  be  transferred 
with  the  dividers.)  Pass  a  circle  about  points  A,  B,  and 
C.  This  circle  is  equal  to  small  circle  XY  of  the  sphere  in 
Fig.  125,  No.  1. 

II.  Find  diameter  XY  of  the  small  circle.  Draw  (Fig.  125,  No.  3) 
a  segment  XY.  With  X  and  Y  as  centers  and  XP  (Fig.  125, 
No.  1)  as  radius  locate  point  P.  Pass  a  circle  about  points 
X,  Y,  and  P.  This  circle  is  equal  to  a  great  circle  of  the 
sphere,  and  its  diameter  is  equal  to  the  diameter  of  the 
sphere. 


104  SOLID   GEOMETRY 

7.  The  intersection  of  two  spherical  surfaces  is  a  circle  (Fig.  1 21  >) . 
Analysis  and  construction  (Fig.  126): 

I.  To  prove  that  the  intersection  of  spheres 
A  and  B  is  a  circle,  prove  that  the  in- 
tersection may  be  generated  by  the 
revolution  of  a  straight  line  that  is 
bisected  at  right  angles  by  the  line  of 
centers  of  the  two  spheres. 

II.  .'.  pass  any  plane  through  the  line  of  centers  of  the  two  spheres. 
This  plane  will  cut  the  spheres  in  great  circles  that  intersect 
in  two  points  C  and  D.  Prove  AB  a  _L  bisector  of  (7). 
(See  Plane  Geometry,  Ths.  72  and  73.) 

8.  If  two  spheres  intersect,  the  line  of  centers  meets  the  spheres 
in   points   which    are  poles   of  the   common  circle   of  the   two 
spheres.  • 

9.  The  locus  of  points  of  contact  of  lines  tangent  to  a  given 
sphere  from  a  given  point  without  the  sphere  is  a  circle. 

Suggestion.  Show  that  this  locus  may  be  regarded  as  the  inter- 
section of  two  spheres. 

10.  The  locus  of  lines  tangent  to  a  sphere  from  a  given  point 
without  the  sphere  is  a  conical  surface. 

NOTE.  If  the  elements  of  a  conical  surface  are  tangent  to  a  sphere, 
the  conical  surface  is  said  to  be  tangent  to  the  sphere  and  is  circum- 
scribed about  the  sphere. 

Suppose  (Fig.  127)  that  a  conical  sur- 
face is  tangent  to  a  given  sphere  and  that 
the  vertex  of  the  conical  surface  recedes 
from  the  sphere.  What  change  will  take 
place  in  the  circle  in  which  the  elements 
of  the  cone  are  tangent  to  the  sphere? 

What  is  the  limiting  position  of  this  circle?     What  is  the  limiting  form 
of  the  cone?     From  these  considerations  we  may  infer  that 

(1)  The  locus  of  the  points  of  contact  of  parallel  tangents  to  a 
sphere  is  a  great  circle. 

(2)  The  locus  of  parallel  tangents  to  a  sphere  is  a  cylindrical  surface. 
If  the  elements  of  a  cylindrical  surface  are  tangent  to  a  sphere, 

the  cylindrical  surface  is  said  to  be  tangent  to  the   sphere  and  is 
circumscribed  about  the  sphere. 

11.  Find  the  locus  of  centers  of  spheres  that  pass  through 
'(1)  two  given  points;    (2)  three  given  points. 


THE   SPHERE  105 

12.  Find  the  locus  of  centers  of  spheres  that  shall  be  tangent 
to  (1)  a  given  line  at  a  given  point;  (2)  a  given  plane  at  a  given 
point. 

EXERCISES   ANALOGOUS   TO    CERTAIN   PLANE 
GEOMETRY   THEOREMS 

126.  NOTE.  Many  plane  geometry  theorems  are  true  also  on  the 
surface  of  the  sphere  if  we  replace  the  straight  line  by  the  great  circle 
arc.  No  plane  geometry  theorems  that  depend  in  any  way  upon 
parallels  can  be  applied  to  the  sphere.  The  following  exercises  are 
illustrations  of  some  of  the  theorems  that  are  true  for  both  plane  and 
spherical  geometry.  It  is  well  for  the  teacher  to  assign  these  exercises 
in  the  given  order,  as  the  later  ones  depend  in  some  cases  on  the  earlier 
ones. 

1.  Give  the  plane  geometry  definitions  for  each  of  the  follow- 
ing terms,  and  if  possible  give  the  definitions  for  the  corresponding 
terms  in  spherical  geometry:  adjacent  angles,  right  angles,  vertical 
angles,  supplementary  angles,  complementary  angles. 

2.  Only  one  great  circle  can  be  drawn  through  a  given  point 
perpendicular  to  another  great  circle,  unless  the  given  point  is 
the  pole  of  the  second  great  circle.     In  this  latter  case  how  many 
great  circle  arcs  can  be  drawn? 

3.  Construct  an  arc  of  a  great  circle  that  is  a  perpendicular 
bisector  of  a  given  great  circle  arc. 

4.  Construct  the  great  circle  arc  that  bisects  a  given  spherical 
angle. 

5.  Construct  a  circle  circumscribed  about  a  given  spherical 
triangle. 

6.  The  great  circle  arc  that  bisects  the  vertex  angle  of  an 
isosceles  spherical  triangle  is  perpendicular  to  the  base  and  bisects 
the  base. 

7.  The  great  circle  arc  that  passes  through  the  vertex  and  the 
mid-point  of  the  base  of  an  isosceles  spherical  triangle  is  perpen- 
dicular to  the  base  and  bisects  the  vertex  angle. 

8.  Are  plane  geometry   Theorems  54   and   55   true  for   the 
spherical  surface?     If  so,  give  proof. 

8 


CHAPTER   IV 

AREAS  AND  VOLUMES 

AREAS   OF  POLYHEDRONS 

GENERAL    STATEMENT 

127.  The  area  of  any  polyhedron  is  the  sum  of  the  areas 
of  its  faces. 

Exercise.     Find  the  total  area  of  a  regular  tetrahedron  if  each 
side  is  4  in.,  6  in.;  if  each  side  is  s. 

THE   LATERAL   AREA   OF  PRISMS 

128.  THEOREM  96.    The  lateral  area  of  any  prism  is 
the  product  of  the  perimeter  of  a  right  section  and  a  lateral 
edge. 


FIG.  128 

Hypothesis:  P  is  any  prism.  R  is  its  right  section.  FI, 
F2,  etc.,  are  its  faces.  e\,  e2,  etc.,  are  its  edges.  ai-f-a2+etc. 
is  the  perimeter  of  R. 

Conclusion:     The  lateral  area  of  P  is  0(01+02+ etc.). 

Analysis: 
I.  To  find  the  lateral  area  of  P,  find  the  area  of  each  of 

the  lateral  faces  and  add. 

II.  To  find  the  area  of  Flt  use  e^  as  the  base  and  prove 
that  ai  is  the  altitude. 
106 


AREAS  AND  VOLUMES 


Proof: 

STATEMENTS 

I.  ai,  02,  etc.,  are  the  altitudes  of  FI,  F2,  etc.,  respec- 

tively. 
II.  e\  —  ez  =  €3  =  etc. 

III.  /.  area  F\= 

area  F^  — 
area  Fz  = 
....  etc. 

IV.  Adding,  the  sum  of  the  areas  of  the  lateral  faces  is 


Let  the  pupil  give  all  reasons.  In  IV  use:  The  sum  of  numbers 
having  a  common  factor  is  the  common  factor  multiplied  by  the  sum 
of  the  coefficients.  How  does  this  theorem  apply? 

COR.  The  lateral  area  of  a  right  prism  is  the  product 
of  the  perimeter  of  the  base  and  a  lateral  edge. 

THE  LATERAL   AREA   OF   A  REGULAR   PYRAMID 

129.  THEOREM  97.  The  lateral  area  of  a  regular  pyramid 
is  one-half  the  product  of  the  perimeter  of  its  base  and  the 
slant  height. 


FIG.  129 

Analysis:  To  find  the  lateral  area  of  P,  find  the  sum  of 
the  areas  of  the  lateral  faces.  In  the  figure  FI,  F2,  etc.,  are 
the  faces  of  the  pyramid.  eit  <?2,  etc.,  are  the  edges  of  the 
base,  ai,  a2,  etc.;  are  the  altitudes  of  the  lateral  faces. 

Let  the  pupil  give  the  proof.  Use  the  proof  of  Th.  96  as  a  model. 
The  slant  height  is  the  common  factor  in  adding. 


108  SOLID   GEOMETRY 

THE  LATERAL   AREA   OF   A  FRUSTUM    OF   A 
REGULAR   PYRAMID 

130.  THEOREM  98.  The  lateral  area  of  a  frustum  of  a 
regular  pyramid  is  one-half  the  product  of  its  slant  height 
and  the  sum  of  the  perimeters  of  its  bases. 


FIG.  130 

Analysis:     To  find  the  lateral  area  of  the  frustum,  find 
the  area  of  each  of  the  lateral  faces  and  add. 

EXERCISES  INVOLVING  AREAS  OF  POLYHEDRONS 

131.  1.  Find  the  total  area  of  a  regular  triangular  prism  if  one 
side  of  its  base  is  3  in.  and  the  lateral  edge  is  5  in. 

2.  Find  the  total  area  of  a  regular  hexagonal  prism  if  one  side 
of  the  base  is  2  in.  and  a  lateral  edge  is  4  in. 

3.  What  will  be  the  cost  at  10 p  a  square  yard  of  painting  the 
lateral  surface  of  a  tower  in  the  form  of  a  regular  octagonal  prism 
if  each  side  of  the  base  is  5  ft.  and  the  height  is  25  ft.? 

4.  The  total   area   of   a   rectangular  parallelepiped  is  82  sq. 
in.     Two  dimensions  are  7  in.  and  2  in.     Find  the  third  dimension. 

5.  Find  the  lateral  area  of  a  regular  octagonal  pyramid  if  the 
slant  height  is  15  in.  and  one  edge  of  the  base  is  7  in.     Find  the 
cost  of  gilding  the  same  at  5<?  a  square  inch. 

6.  What  is  the  cost  at  15?  a  square  foot  of  painting  a  steeple 
in  the  form  of  a  regular  square  pyramid  if  one  side  of  the  base  is 
9  ft.  and  the  height  is  40  ft.? 

7.  Find  the  total  area  of  a  regular  square  pyramid  if  one  edge 
of  the  base  is  10  in.  and  one  lateral  edge  13  in;  if  one  edge  of  the 
the  base  is  a  and  one  lateral  edge  e. 


AREAS  AND  VOLUMES  100 

8.  Find  the  total  area  of  a  regular  tetrahedron  whose  edge 
is  10  in. 

9.  Find  the  total  area  of  a  regular  octahedron  whose  edge  is 
7  in. 

10.  Find  the  total  area  of  a  frustum  of  a  regular  hexagonal 
pyramid  if  the  sides  of  the  bases  are  2  ft.  and  1  ft.  10  in.  respectively 
and  a  lateral  edge  is  4  ft. 

VOLUMES   OF   POLYHEDRONS 
MEASURING    SPACE 

132.  To  measure  the  space  inclosed  by  the  surface  of  a 
polyhedron  is  to  find  how  many  times  it  contains  another 
solid  used  as  a  unit  of  measure. 

The  volume  of  a  polyhedron  is  the  measure  number  of 
the  space  inclosed  by  the  surface  of  the  polyhedron. 

While  any  solid  might  be  used  as  a  unit  of  volume,  it  is 
the  common  practice  to  use  a  cube  whose  edge  is  a  unit  of 
length.  Any  segment  may  be  used  as  a  unit  of  length  with 
the  corresponding  units  of  area  and  of  volume.  It  is,  how- 
ever, most  convenient  practically  to  use  one  of  the  recog- 
nized standard  units  of  length;  thus  we  have  one  inch,  one 
square  inch,  one  cubic  inch;  one  foot,  one  square  foot,  one 
cubic  foot;  one  centimeter,  one  square  centimeter,  one  cubic 
centimeter,  etc. 

FUNDAMENTAL   ASSUMPTION 

133.  As.  27.  The  number  of  units  of  volume  in  a  rec- 
tangular parallelepiped  is  the  product  of  the  number  of 
linear  units  in  three  edges  that  meet  at  a  common  vertex. 

If  V  represents  the  volume  of  the  rectangular  parallele- 
piped, and  a,  6,  and  c  the  length  of  three  edges  that  meet 
at  a  common  vertex,  As.  27  may  be  stated  as  a  formula: 

V  =  abc 


110 


SOLID    GEOMETRY 


The  assumption  will  be  discussed  under  two  heads: 

134.  A.    When  the  edges  of  the  rectangular  parallele- 
piped are  all  commensurable  with  a  given  unit  of  length. 

In  this  case  the  unit 
of  length  can  be  applied 
an  integral  number  of 
times  to  each  edge.  Sup- 
pose the  unit  of  length 
is  contained  in  one  side 
of  the  base  a  times,  and  


in  the  adjacent  side  of 

the  base  b  times;  then  the  base  of  the  rectangular  paral- 
lelepiped can  be  divided  into  ab  unit  squares.  The  unit  of 
length  may  be  contained  in  the  height  of  the  parallelepiped 
c  times.  By  planes  passing  through  the  points  of  division, 
the  parallelepiped  may  be  divided  into  c  layers  with  ab 
unit  cubes  in  a  layer.  The  volume  is,  therefore,  abc  units 
of  volume  (Fig.  131). 

Suppose  one  or  more  edges  of  the  parallelepiped  are  not 
exactly  divisible  by  the  unit  chosen,  but  are  divisible  by 
some  aliquot  part  of  the  unit.  In  this  case  this  part  of  the 
chosen  unit  may  be  taken  as  a  new  unit  of  length,  and  a 
cube  whose  edge  is  this  new  linear  unit  may  be  considered 
as  the  unit  of  volume.  The  assumption  is  then  evident  as 
above. 

Let  the  pupil  give  special  cases  as  illustrations. 

135.  B.  When  one  or  more  edges  of  the  rectangular 
parallelepiped  are  incommensurable  with  the  chosen  unit. 

In  this  case  it  is  not  possible  to  measure  one  side  or  per- 
haps all  sides  of  the  parallelepiped  in  integral  or  fractional 
terms  of  the  chosen  unit.  Since  the  ratio  of  two  incom- 
mensurable segments  is  an  irrational  number,  these  sides 
may  be  expressed  in  irrational  terms  of  the  chosen  unit  and 
the  volume  found.  This  volume  may  be  irrational  or 
rational  according  to  the  measure  number  of  the  sides. 


AREAS  AND  VOLUMES  111 

The  following  illustrations  come  under  this  case: 

1.  Length  incommensurable  expressed  as  an  irrational 
number : 

Length  3V27width  3,  height  5,  volume  45>/2~ 

2.  Length  and  width  incommensurable: 

a.  Length  V27  width  2Vs7  height  5,  volume  10V(T 

b.  Length  2V2,"width  3>/27 height  5,  volume  60 

Let  the  pupil  give  illustrations  of  cases  in  which  all  three  dimen- 
sions are  expressed  by  irrational  numbers. 

Since  an  irrational  number  cannot  be  expressed  exactly 
as  an  integer  or  a  fraction,  rectangular  parallelepipeds  such 
as  those  mentioned  in  the  illustrations  above  cannot  be 
divided  into  unit  cubes  by  planes  passing  through  points 
of  division  as  in  Case  A.  These  dimensions  can,  however, 
be  expressed  as  approximate  decimals,  and  these  approxi- 
mations may  be  made  as  close  as  we  choose  to  make  them 
to  the  true  dimensions.  From  the  approximate  dimensions 
an  approximate  volume  may  be  computed,  and  this  approxi- 
mate volume  may  be  as  close  as  we  choose  to  make  it  to  the 
true  volume. 

Illustration.     If   the   length  is   V27  the  width   VST  the 
height  5,  we  may  have  the  following  approximations: 
Length  1.4,  width  1.7,  height  5,  volume  12 
Length  1.41,  width  1.73,  height  5,  volume  12.2 
Length  1.414,  width  1.732,  height  5,  volume  12.24 

This  process  may  be  continued  indefinitely,  as  the  deci- 
mals expressing  V2  and  V3  do  not  terminate. 

Ex.  1.  Find  the  approximate  volume  of  a  rectangular  parallele- 
piped correct  to  two  decimal  places,  if  its  dimensions  are  V3, 
2V27  and  VsT  Find  the  total  area. 

Ex.  2.  What  is  the  answer  to  Ex.  1  if  the  dimensions  of  the 
parallelepiped  are  V3  ,  2\/2  ,  and  V(T?  Find  the  total  area. 

Ex.  3.  Find  the  total  area  and  the  volume  of  a  rectangular 
parallelepiped  whose  dimensions  are  2%  in.,  5H  in.,  'and  3%  in. 


112 


SOLID   GEOMETRY 


THE   VOLUME    OF   ANY   PARALLELEPIPED 

136.  NOTE.     If  the  teacher  desires,  the  treatment  of  volumes  by 
Cavalieri's  theorem  (§§173-175)  maybe  substituted  for  §§142  and  144. 
In  this  case  §§136-141  may  be  omitted. 

Two  solids  that  fill  the  same  extent  of  space  are  said  to 
be  equivalent.  Congruent  solids  are  the  simplest  examples 
of  equivalent  solids. 

137.  THEOREM  99.     Two  truncated  right  prisms  are  con- 
gruent if  their  right  bases  are  congruent  and  three  lateral 
edges  of  one  are  equal  respectively  to  three  corresponding 
lateral  edges  of  the  other,  and  similarly  placed. 


C' 


FIG.  132 

Hypothesis:     P  and  P'  are  two  truncated  right  prisms 
with  the  right  bases  b  and  b'  congruent,  and  the  edges 

A'F't  BG  =  B'Gr,  and  CH  =  C'H'. 
Conclusion:     P  ^  P'. 
Outline  of  proof: 

L  Base  b  can  be  made  to  coincide  with  base  b'. 
II.  The  lateral  edges  of  P  will  fall  along  the  correspond- 
ing lateral  edges  of  P'. 

III.  The  points  F,  G,  and  H  will  fall  on  the  points  F', 

G't  and  Hf  respectively. 

IV.  The  plane  of  the  upper  base  of  P  will  fall  on  the 

plane  of  the  upper  base  of  P'. 

V.  Points  7,  /,  etc.,  will  fall  on  points  /',  ]'  respectively. 
VI.  P  and  P'  will   coincide  and  be  congruent. 
Is  it  necessary  for  the  given  lateral  edges  to  be  consecutive? 


AREAS  AND   VOLUMES 


113 


COR.  Two  right  prisms  are  congruent  if  they  have 
congruent  bases  and  equal  altitudes. 

138.  THEOREM  100.  If  an  oblique  prism  and  a  right 
prism  are  cut  from  the  same  prismatic  space  and  have 
equal  lateral  edges,  they  are  equivalent. 


A 


A' 


A 


7 


FIG.  133 


FIG.  134 


Hypothesis:  R  is  a  right  prism  and  O  an  oblique  prism 
cut  from  the  same  prismatic  space.  The  edge  AB  equals 
the  edge  A  'B'  (Fig.  133). 

Conclusion :     R  =  0. 

Analysis:  To  prove  R  =  O,  prove  the  truncated  right 
prisms  R+T  and  T+0  congruent  and  subtract  from  each 
the  truncated  prism  T. 

Discussion:  A  complete  proof  of  Th.  100  requires  a 
discussion  of  two  cases. 

Case  A.     When  the  prisms  are  not  telescoped. 

Case  B.     When  the  prisms  are  telescoped. 

Let  the  pupil  give  the  proof  for  Case  B,  using  Fig.  134.  Notice 
that  two  solids  are  equivalent  if  they  are  sums,  differences,  or  equal 
parts  of  equivalent  solids. 

COR.  Two  prisms  having  equal  edges  and  congruent 
right  sections  are  equivalent. 


114 


SOLID   GEOMETRY 


139.  THEOREM  101.    The  volume  of  any  parallelepiped 
is  the  product  of  the  area  of  its  base  and  its  altitude. 


FIG.  135 

Hypothesis:    P  is  any  parallelepiped,  b  is  its  base,  and  h 
its  altitude. 
.  Conclusion:    The  volume  of  P  =  bh. 

Analysis  in  general:  To  prove  that  the  volume  of  P  =  bh, 
compare  P  with  a  rectangular  parallelepiped  that  has  a  base 
equivalent  to  b  and  an  altitude  equal  to  h. 

Construction: 

I.  Extend  the  edges  of  P  that  are  parallel  to  BA  and 

take  B'A'=BA.  Pass  planes  A'H'  and  B'G'  _L 
A  'B'  at  A'  and  B'  respectively.  This  gives  paral- 
lelepiped PI. 

II.  Extend  the  edges  of  Pl  that  are  parallel  to  B'C,  and 

take  B"C"  =  B'Cr.  Pass  planes  B"E"  and  C"H" 
_L  B"C"  at  B"  and  C"  respectively.  This  gives 
parallelepiped  P^ 

Outline  of  proof: 

It  is  necessary  to  prove: 
I.  P  =  P2. 

II.  PZ  is  a  rectangular  parallelepiped. 

III.  b"  =  ba.ndk  =  h. 

.'.  since  vol.  P2  =  6"fc,  vol.  P  =  b"k;  but  b"  =  b  and  k  =  h. 
:.  vol.  P  =  bh. 


AREAS   AND   VOLUMES  115 

Analysis    for   I:     To   prove   P  =  P2,    prove   P  =  Pi  =  P2. 
(Use  Th.  100.) 
Analysis  for  II: 

a.  To  prove  P2  rectangular,  prove  that  the  edges  B"A", 

B"C",  and  B"F"  are  each  _L  the  other  two. 

b.  To  prove  B"F"  _L  A"B"  and  B"C",  prove  B"F" 

J_  plane  of  b" . 

c.  :.  prove  planes  B"C"G"F"  and  B"A"E"F"  each  _L 

A"5"C"£>".     (See  Ths.  25  and  28.     Use  the  con- 
struction.) 

Analysis  for  III:  Prove  b  =  b'  =  b".  (See  Plane  Geom- 
etry, Th.  116.) 

THE   VOLUME   OF   TRIANGULAR   PRISMS 

140.  THEOREM  102.  A  plane  passed  through  two  diago- 
nally opposite  edges  of  a  parallelepiped  divides  the  paral- 
lelepiped into  two  equivalent  triangular  prisms. 

7* 


FIG.  136 
Analysis: 

I.  To  prove  T\  —  T^  prove  that  they  have  equal  edges 
and  congruent  right  sections. 

II.  /.  construct  the  right  section  WXYZ  and  prove  that 

(1)  The  edges  of  T\  are  equal  to  the  edges  of  TV 

(2)  The  right  section  WXZ  of  TI  is  congruent  to 
the  right  section  XYZ  of  T2. 

III.  .*.  prove  WXYZ  a  parallelogram. 

Are  TI  and  T2  congruent?     Illustrate  with  a  model. 


116 


SOLID   GEOMETRY 


141.  THEOREM  103.     The  volume  of  a  triangular  prism 
is  the  product  of  the  area  of  its  base  and  its  altitude. 


FIG.  137 

Hypothesis:  T  is  a  triangular  prism  with  base  b  (ABC) 
and  altitude  h. 

Conclusion:     The  volume  of  T  =  bh. 

Analysis  in  general:  To  find  the  volume  of  T,  compare 
it  with  a  parallelepiped. 

Construction:  Complete  the  parallelogram  that  has  A  B 
and  BC  for  sides.  Complete  the  parallelogram  that  has  EF 
and  FG  for  sides.  Join  HD. 

Outline  of  proof: 

I.  ABCD-EFGH  is  a  parallelepiped  (P). 
II.  Volume  of  T=H  volume  of  P. 

III.  Volume  of  P  =  ABCD  •  h. 

IV.  /.  volume  of  T  =  MABCD-h. 
V.  AABC  =  y2ABCD  =  b. 

VI.    .'.  volume  of  T  =  bh. 

Analysis  for  I: 

a.  To  prove  ABCD-EFGH  a  parallelepiped,  prove  HD  \\ 

AE. 

b.  :.  prove  AEHD  SL  O: 

Exercise.  In  Fig.  137,  construct  the  parallelepiped  by  passing 
planes  through  GC  and  AE  parallel  to  planes  ABFE  and  BCGF 
respectively  and  extending  the  planes  of  the  upper  and  lower 
bases.  Prove  that  the  solid  formed  is  a  parallelepiped. 


AREAS  AND   VOLUMES 


117 


THE  VOLUME   OF  ANY  PRISM 

142.  THEOREM  104.     The  volume  of  any  prism  is  the 
product  of  the  area  of  its  base  and  its  altitude. 

j  _ 

i 


FIG.  138 

Analysis  arid  construction:  To  find  the  volume  of  P, 
divide  it  into  triangular  prisms  and  add  their  volumes. 

In  the  proof  use  the  theorem:  The  sum  of  numbers  having  a 
common  factor  is  the  common  factor  multiplied  by  the  sum  of  the 
coefficients. 

THEOREM  105.  If  two  prisms  have  equivalent  bases  and 
equal  altitudes,  they  are  equivalent. 


EXERCISES  INVOLVING  VOLUMES  OF  PRISMS 

143.  1.  Find  the  volume  of  a  regular  triangular  prism  if  one 
side  of  the  base  is  15  in.  and  the  height  is  10  ft. 

2.  The  corner  of  a  cellar  is  boarded  off  to  form  a  triangular 
coal  bin.     How  many  tons  will  it  hold  if  the  base  is  an  isosceles 
right  triangle  each  leg  of  which  is  8  ft.  and  it  is  4  ft.  deep?     Allow 
35  cu.  ft.  to  one  ton. 

3.  The  base  of  a  triangular  prism  is  an  isosceles  right  triangle 
with  the  hypotenuse  8  in.     If  the  height  of  the  prism  is  15  in., 
find  its  volume. 

4.  Find  the  volume  of  a  regular  hexagonal  prism  if  its  height 
is  10  in.  and  one  side  of  the  base  is  3  in. 

5.  The  base  of  a  parallelepiped  is  a  rhombus  with  one  side 
12  in.  and  one  angle  60°.     Find  its  volume  if  its  height  is  24  in. 


118 


SOLID   GEOMETRY 


6.  A  stick  of  timber  is  8  in.  x  15  in.  and  20  ft.  long.      Find  the 
weight  if  1  cu.  ft.  weighs  50  Ibs. 

7.  The  space  left  in  the  basement  for  a  coal  bin  is  10  ft.  X 12  ft. 
How  deep  must  the  bin  be  to  hold  10  tons? 

8.  The  area  of  a  cube  is  96  sq.   in.     Find  its  volume  and 
its  diagonal. 


n 


THE   VOLUME   OF   A   TRIANGULAR   PYRAMID 

144.  In  Fig.  139,  the  altitude  of  pyramid  P  is  divided  into 
any  number  of  equal  parts.  Through  the  points  of  division 
planes  are  passed  parallel  to 
the  base  of  the  pyramid. 
These  planes  cut  the  pyramid 
the  sections  AiBiCi, 
,  etc.  On  these  sections 
as  upper  bases  prisms  are 
constructed  by  planes  passing 
through  the  lines  Bid,  BZC2, 
etc.,  parallel  to  the  edge  AP. 
The  series  of  prisms  thus  FlG-  139 

formed  is  said  to  be  inscribed  in  the  pyramid.  Show  that 
the  edges  of  these  prisms  are  parallel  to  the  edge  PA  of 
the  pyramid. 

It  is  evident  that  if  we 
increase  the  number  of  divi- 
sions in  the  altitude  we 
shall  increase  also  the  num- 
ber of  inscribed  prisms,  and 
that  the  number  of  inscribed 
prisms  may  be  increased  in- 
definitely (Fig.  140). 

It  is  evident  also  that  if  a    __  _ 
series  of  prisms  is  inscribed  FlG-  14U 

in  a  pyramid,  and  if  their  number  is  increased  indefinitely, 


AREAS  AND  VOLUMES 


119 


a   series  of  prisms  will  soon  be  obtained  that  can  with 
difficulty  be  distinguished  from  the  pyramid. 

The  following  theorem  can  be  proved  in  higher  mathe- 
matics and  will  be  assumed  here : 

As.  28.  There  is  a  definite  limit  to  the  sum  of  the  vol- 
umes of  a  series  of  prisms  inscribed  in  a  given  pyramid  if 
their  number  is  increased  indefinitely. 

This  limit  is  by  definition  the  volume  of  the  pyramid. 


FIG.   141 

Fig.  141  shows  two  pyramids,  P  and  P',  standing  on 
the  same  plane.  They  have  equal  altitudes  and  equivalent 
bases.  Suppose  that  the  altitude  h  is  divided  into  equal 
parts,  and  that  through  the  points  of  division  planes  are 
passed  cutting  both  pyramids,  and  that  on  these  sections  as 
upper  bases  a  series  of  prisms  is  inscribed  in  each  pyramid. 

Prove  that  the  sum  of  the  volumes  of  the  prisms  in  P  is 
always  equal  to  the  sum  of  the  volumes  of  the  prisms  in  P'. 

It  can  be  proved  in  higher  mathematics  that,  since  the 
sum  of  the  volumes  of  the  prisms  of  the  series  in  P  is  always 
equal  to  the  sum  of  the  volumes  of  the  prisms  in  P't  the 
limits  of  these  two  sums  are  equal. 

We  will,  therefore,  assume 

THEOREM  106.  If  two  pyramids  have  equivalent  bases 
and  equal  altitudes,  they  are  equivalent. 


120 


vSOLID   GEOMETRY 


145.  THEOREM     107.     Any    triangular    prism    may 
divided  into  three  equivalent  triangular  pyramids. 


be 


FIG.  142 


Hypothesis:     P  is  a  triangular  prism  cut  by  the  planes 
XBC  and  XYC,  forming  three  triangular  pyramids. 
Conclusion:     X-A  BC  =  C-XYZ  =  X-CBY . 
Analysis: 
I.  To    prove    pyramid     I  =  pyramid     II,     prove    base 

ABC  =  base   XYZ,   and   their   altitudes   equal. 
II.  To  prove  pyramid  11=  pyramid  III,  regard  X  as  the 

vertex  of  each,  and  CYZ  and  CBY  as  bases. 
146.  THEOREM  108.     The  volume  of  a  triangular  pyramid 
is  one-third  the  product  of  the  area  of  its  base  and  its 
altitude. 


FIG.  14.°, 


Hypothesis:     D-ABC  is  a  triangular  pyramid  with  base  b 
(ABC)  and  altitude  h. 

Conclusion:     The  volume  of  D-ABC  = 


AREAS  AND  VOLUMES  121 

Analysis  in  general:  To  find  the  volume  of  D—ABC, 
compare  it  with  a  triangular  prism  having  the  same  base 
and  altitude. 

Construction:     From  A  and  C  draw  lines  parallel  to  BD. 
Through  D  pass  a  plane  parallel  to  ABC. 
Outline  of  proof: 

I.  ABC-DEF  is  a  prism. 
II.  Volume  of  D-ABC  =  }/$  volume  of  ABC-DEF. 

III.  Volume  of  ABC-DEF  =  bh. 

IV.  .'.  volume  of  D- 


147.  THEOREM  109.    The  volume  of  any  pyramid  is  one- 
third  the  product  of  the  area  of  its  base  and  its  altitude. 


FK;.  144 

Suggestion.  Divide  P  into  triangular  pyramids,  and  add  their 
volumes. 

GENERAL    FORMULA    FOR    THE    VOLUME    OF    PRISMS, 
PYRAMIDS,   AND    FRUSTUMS 

148.  A  prismatoid  is  a  polyhedron  that  has  for  bases  two 
polygons  in  parallel  planes,  and  for  lateral  faces  triangles, 
trapezoids,  or  parallelograms  that  have  one  side  in  common 
with  one  base  and  the  opposite  vertex  or  side  in  common 
with  the  other  base.  The  altitude  of  a  prismatoid  is  the  per- 
pendicular distance  between  the  bases.  The  mid-section  of 
a  prismatoid  is  a  section  made  by  a  plane  parallel  to  the 
bases  and  bisecting  the  altitude. 

9 


122 


SOLID   GEOMETRY 


THEOREM  110.  The  volume  of  a  prismatoid  is  one-sixth 
the  altitude  times  the  sum  of  the  areas  of  the  bases  and 
four  times  the  area  of  the  mid-section. 

X 


FIG.  145 

Hypothesis:  P  is  a  prismatoid  with  b  the  lower  base,  b' 
the  upper  base,  m  the  mid-section,  h  the  altitude,  and  V  the 
volume. 

Conclusion:     V  =  Xh(b+b'+4m). 

Analysis:  To  find  V,  divide  P  into  triangular  pyramids 
and  add  their  volumes. 

Construction:  Join  0  any  point  in  m  with  the  vertices 
of  b  and  &'.  Draw  one  diagonal  in  each  of  these  lateral  faces 
which  are  trapezoids  and  parallelograms.  0  is  the  common 
vertex  of  a  series  of  triangular  pyramids  and  also  of  those 
pyramids  whose  bases  are  the  upper  and  the  lower  bases  of 
the  prismatoid. 

Outline  of  proof: 

I.  Vol.  of  0-XYZ  =  X-)4h-b'  =  Xhb'. 
II.  Vol.  of  0-ABC  etc.  =  Y*  •  Xh  -  b  =  Hhb. 
III.  Vol.  of  0-XAB  =XAB  •  H  altitude  from  0  to  XAB. 
R  and  5  are  mid-points  of  AX  and  BX  respec- 
tively. 

AABX  =  ±  &RSX. 
.'.  vol.  0-XAB  =  4  vol.  of  0-RSX. 

0-RSX=X-ORS. 
:.  vol.  0-XAB  =  4  •  H  •  Mfc  •  ORS  =  Xh  -  4  ORS. 


AREAS  AND   VOLUMES  123 

IV.  The  sum  of  the  volumes  of  these  pyramids  that  have 
0  as  a  vertex  and  the  lateral  faces  as  bases  is 
l/Qh  •  4  m. 

V.    /.  the  sum  of  the  volumes  of  all  pyramids  is 
Xh(b+b'+4m) 

149.  Prisms,  pyramids,  and  frustums  of  pyramids  are 
special  cases  of  prismatoids.  How?  Their  volumes  may 
be  obtained  from  the  volume  of  the  prismatoid  as  shown 
below.  In  each  case  it  is  necessary  to  find  the  volume  in 
terms  of  b  and  b'. 

COR.  I.    The  volume  of  any  prism  is  bh. 

Suggestion.     b  =  b'  =  m.     Then 


COR.  II.    The  volume  of  any  pyramid  is 

Suggestion.     b'=0,m  =  }4b.     Then 


COR.  III.    The  volume  of  a  frustum  of  a  pyramid  is 


Outline  of  proof  (Fig.  146)  : 

b     a2       AV    a'2 
—  =  -5  and  —  =  —  ;  • 
m    %*          m     %z 

V~6     a         .   VF     a' 
/.-==  =  -  and—  F^  =  —  • 
x  Vm     x 


Vfr  +V&/=q+o/  =  2 

Vw  x  

,  7  /  FIG.  146 

—  =  4. 


Since  V±Mh(b+b'+4m),  where  /i  is  altitude. 


124  SOLID   GEOMETRY 

EXERCISES    INVOLVING    VOLUMES    OF    POLYHEDRONS 

160.  1.  A  grain  bin  is  2^  ft.  wide,  6  ft.  long,  and  4  ft.  deep. 
How  many  bushels  of  grain  will  it  hold?     2150.42  cu.  in.  =  l  bu. 

2.  What  would  be  the  area  of  the  base  of  a  grain  bin  6 .ft.  deep 
to  hold  250  bu.? 

3.  The  volume  of  any  prism  is  the  product  of  the  area  of  a 
right  section  and  a  lateral  edge. 

4.  The  lateral  edge  of  a  triangular  prism  is  15  in.     The  right 
section  is  an  equilateral  triangle  5  in.  on  a  side.     Find  the  volume. 

5.  If  the  dimensions  of  a  rectangular  parallelepiped  are  a,  b, 
and  c,  write  a  formula  for  (1)  the  sum  of  the  edges ;  (2)  the  diagonal ; 
(3)  the  total  area;   (4)  the  volume. 

6.  When  the  rainfall  is  y%  in.,  how  many  barrels  of  water  will 
fall  per  acre?     1  cu.  ft.  of  water  =7^  gal.;  313/2  gal.  =  1  bbl. 

7.  The  plane  determined  by  one  edge  of  a  tetrahedron  and 
the  mid-point  of  the  opposite  edge  divides  the  pyramid  into  two 
equivalent  parts. 

8.  Lines  drawn  from  the  center  of  a  cube  to  the  vertices  divide 
it  into  six  equivalent  pyramids. 

9.  Find  the  volumes  of  the  following  pyramids  if 

a.  Base  is  a  regular  hexagon  5  cm.  on  a  side  and  the  altitude 
is  9  cm. 

b.  Base  is  a  rectangle  10  in.  Xl4  in.  and  the  altitude  is  18  in. 

10.  Find  the  volume  of  a  regular  octahedron  8  in.  on  a  side. 

1 1 .  Find  the  total  area  and  the  volume  of  a  regular  hexagonal 
pyramid  if  one  edge  of  the  base  is  8  in.  and  one  lateral  edge  is 
10  in. 

12.  Find  a  formula  for  the  volume  of  a  pyramid  whose  base 
is  an  equilateral  triangle  with  one  side  e  and  the  altitude  h. 

13.  What  is  the  formula  called  for  in  the  previous  exercise  if 
the  base  is  (1)  a  square;  (2)  a  regular  hexagon? 

14.  The  edges  of  the  bases  of  a  frustum  are  15  in.  and  9  in. 
respectively,  and  its  height  is  4  ft.     Find  its  volume  if  it  is  a  frus- 
tum of  (1)   a  regular  square  pyramid;   (2)   a  regular  triangular 
pyramid. 

15.  Find  the  volume  of  the  pyramids  from  which  the  frustums 
mentioned  in  the  previous  exercise  are  cut. 


AREAS  AND   VOLUMES  125 

THE   MEASUREMENT   OF  ROUND   BODIES 
IN   GENERAL 

151.  NOTE.  Rigorous  proofs  for  the  theorems  concerning  the 
measurement  of  round  bodies  are  too  difficult  for  this  book.  The 
following  treatment  will  make  the  theorems  concerning  the  measure-' 
ment  of  cylinders  and  cones  appear  reasonable  to  the  pupil.  If  the 
teacher  desires,  the  proofs  of  theorems  concerning  volumes,  based  on 
Cavalieri's  theorem  (§§173,  177,  178),  may  be  substituted  for  §§156 
and  163. 

The  surface  of  a  polygon  that  can  be  drawn  on  a  plane 
may  be  measured  by  finding  how  many  times  it  would  con- 
tain another  polygon  used  as  a  unit  of  measure.  It  is  evi- 
dent that  the  measure  of  a  curved  surface  like  the  lateral 
surface  of  a  cylinder  or  of  a  cone  cannot  be  found  in  this  way. 
We  will  assume,  however,  that  the  area  of  curved  surfaces 
can  be  expressed  in  terms  of  plane  units,  but  it  will  be  neces- 
sary for  us  to  define  what  is  meant  by  the  area  of  these  sur- 
faces. These  definitions  will  be  given  later. 

Similarly,  we  will  assume  that  the  measure  of  the  space 
inclosed  by  cylinders  and  cones  can  be  expressed  in  terms  of 
unit  cubes,  and  will  define  later  what  is  meant  by  the  volume 
of  these  solids. 

THE   MEASUREMENT   OF  THE   CYLINDER 
INSCRIBED    PRISMS 

152.  A  polygon  is  said  to  be  inscribed  in  any  curve  if 
its  vertices  lie  on  the  curve. 

If  the  bases  of  a  prism  are  in- 
scribed in  the  bases  of  a  cylinder, 
and  the  lateral  edges  of  the 
prism  are  elements  of  the  cylin- 
der, the  prism  is  said  to  be  in- 
scribed in  the  cylinder  and  the 

cylinder  is  said  to  be  circum- 

scribed  about  the  prism  (Fig.  147) .  FIG.  147 


126  SOLID   GEOMETRY 

We  will  also  assume 

As.  29.  Any  plane  that  cuts  all  of  the  elements  of  a 
cylinder  will  cut  any  inscribed  prism  in  a  section  which  is 
inscribed  in  the  section  of  the  cylinder. 

Fig.  148  shows  a  triangular  prism  inscribed  in  a  circular 
cylinder.  AD  represents  the  plane  that  cuts  the  cylinder 
and  the  prism  in  right  sec- 
tions. ACE  is  the  right 
section  of  the  prism.  If  the 
arcs  between  the  vertices  of 
A  ACE  are  bisected,  and  the 
points  of  division  are  joined, 
a  hexagon  is  inscribed  in  the 
right  section  of  the  cylinder. 
If  elements  of  the  cylinder  are  - 
drawn  through  the  points  of 

division  A,  B,  C,  D,  E,  and  F,  they  will  form  the  edges  of 
an  inscribed  hexagonal  prism  whose  right  section  ABCDEF 
is  inscribed  in  the  right  section  of  the  cylinder.  It  is  evident 
that  if  this  process  is  continued  the  number  of  lateral  faces 
of  the  inscribed  prism  can  be  increased  indefinitely.  It  is 
to  be  noted  that  this  definition  applies  to  any  closed  cylinder 
and  is  independent  of  the  number  of  lateral  faces  of  the 
original  prism. 

THE   LATERAL   AREA   OF   CIRCULAR   CYLINDERS 

153.  It  is  evident  that  if  a  prism  is  inscribed  in  a  circular 
cylinder,  and  if  the  number  of  lateral  faces  is  increased 
indefinitely,  a  prism  will  soon  be  obtained  that  can  with 
difficulty  be  distinguished  from  the  cylinder. 

The  following  theorem  from  higher  mathematics  will  be 
assumed: 

THEOREM  111.  There  is  a  definite  limit  to  the  lateral 
areas  of  a  series  of  prisms  inscribed  hi  a  circular  cylinder 
when  the  number  of  lateral  faces  is  increased  indefinitely. 


AREAS   AND   VOLUMES  127 

NOTE.  Theorem  111  is  true  no  matter  what  is  the  nature  of  the 
initial  prism  of  the  series. 

The  lateral  area  of  the  cylinder  is  defined  as  the  limit  of 
the  lateral  areas  of  a  series  of  inscribed  prisms  as  the  number 
of  lateral  faces  is  increased  indefinitely. 

Therefore,  since  the  lateral  area  of  a  prism  is  the  prod- 
uct of  the  perimeter  of  a  right  section  and  an  element,  no 
matter  how  many  lateral  faces  the  prism  may  have,  we  will 
assume 

THEOREM  112.  The  lateral  area  of  a  circular  cylinder  is 
the  product  of  the  perimeter  of  a  right  section  and  an  element. 

NOTE.  As  proved  in  higher  mathematics,  Th.  112  is  true  for  any 
cylinder.  It  can  be  used  here  only  for  circular  cylinders,  because  in 
high-school  mathematics  we  do  not  learn  how  to  find  the  perimeters 
of  curves  other  than  circles. 

COR.  I.  If  r  is  the  radius  of  the  right  section,  e  an  ele- 
ment, and  L  the  lateral  area  of  a  circular  cylinder, 


In  Cor.  II,  r  represents  the  radius  of  the  base,  h  repre- 
sents the  height,  L  represents  the  lateral  area,  A  represents 
the  total  area. 


COR.  II.    In  right  circular  cylinders 


(fi  +  r) 


154.  NOTE  1.  Since  the  value  of  TT  can  be  found  only  approximately, 
it  follows  that  the  area  of  circular  cylinders  can  be  found  only  approxi- 
mately. In  brief,  the  lateral  area  of  an  inscribed  prism  of  a  great 
many  lateral  faces  is  taken  as  an  approximation  to  the  lateral  area 
of  the  cylinder. 

165.  NOTE  2.  The  lateral  surface  of  a  right  circular  cylinder  may 
be  developed  or  rolled  out  into  a  rectangle.  Cut  a  rectangular  strip 
of  paper  whose  width  is  equal  to  an  element  of  the  given  cylinder. 
Wrap  the  strip  about  the  cylinder  and  cut  it  with  a  sharp  knife  so  that 
it  exactly  fits  about  the  cylinder  without  overlapping.  Find  its  area. 
Find  the  development  of  the  lateral  surface  of  an  oblique  circular 
cylinder. 


128  SOLID   GEOMETRY 

THE   VOLUME   OF   CYLINDERS   WITH   CIRCULAR   BASES 

166.  In  studying  the  volumes  of  cylinders  we  will  imagine 
that  a  series  of  prisms  is  inscribed  in  a  cylinder  with  a  cir- 
cular base,  and  that  the  number  of  lateral  faces  of  the  prisms. 
is  increased  indefinitely. 

The  following  theorem  from  higher  mathematics  will  be 
assumed  : 

THEOREM  113.  There  is  a  definite  limit  to  the  volumes 
of  a  series  of  prisms  inscribed  in  a  cylinder  with  a  circular 
base  when  the  number  of  lateral  faces  of  the  prisms  is 
increased  indefinitely. 

NOTE.  Theorem  113  is  true  no  matter  what  is  the  nature  of  the 
initial  prism  of  the  series. 

The  volume  of  the  cylinder  is  defined  as  the  limit  of  the 
volumes  of  a  series  of  inscribed  prisms  as  the  number,  of 
lateral  faces  is  increased  indefinitely. 

Therefore,  since  the  volume  of  a  prism  is  the  product  of 
the  area  of  the  base  and  the  altitude,  no  matter  how  many 
lateral  faces  the  prism  may  have,  we  will  assume 

THEOREM  114.  The  volume  of  a  cylinder  with  a  cir- 
cular base  is  the  product  of  the  area  of  its  base  and  its 
altitude. 

NOTE.  As  proved  in  higher  mathematics,  Th.  114  is  true  for  any 
cylinder.  It  can  be  used  here,  however,  only  for  cylinders  with  circu- 
lar bases,  because  in  high-school  mathematics  we  do  not  learn  how  to 
find  the  areas  of  surfaces  inclosed  by  curves  other  than  circles. 

COR.  If  r  is  the  radius  of  the  base  of  a  cylinder,  h  the 
altitude,  and  V  the  volume, 


167.  NOTE  1.  The  volume  of  cylinders  can  be  found  only  approxi- 
mately; in  fact,  the  volume  of  an  inscribed  prism  of  a  great  many 
lateral  faces  is  taken  as  an  approximation  to  the  volume  of  the  cylinder. 

158.  NOTE  2.  By  similar  processes  the  lateral  area  and  the  volume 
of  cylinders  can  be  obtained  from  circumscribed  prisms. 


AREAS  AND  VOLUMES  129 

THE   MEASUREMENT   OF  THE   CONE 

INSCRIBED   PYRAMIDS 

169.  If  the  base  of  a  pyramid  is  inscribed  in  the  base  of 
a  cone,  and  the  lateral  edges  of  the  pyramid  coincide  with 
elements  of  the  cone,  the  pyramid  is  said  to  be  inscribed  in 
the  cone  and  the  cone  is  said  to  be  circumscribed  about  the 
pyramid  (Fig.  149). 

to 


FIG.  149  FIG.  150 

Fig.  150  shows  a  triangular  pyramid  0-ACE  inscribed  in 
a  cone.  If  the  arcs  between  the  vertices  of  &ACE  are 
bisected,  and  the  points  of  division  are  joined,  a  hexagon  is 
inscribed  in  the  base  of  the  cone.  If  elements  of  the  cone 
are  drawn  through  the  vertices  of  this  hexagon,  they  will 
form  the  edges  of  an  inscribed  hexagonal  pyramid.  It.  is 
evident  that  if  this  process  is  continued  the  number  of 
lateral  faces  of  the  pyramid  may  be  increased  indefinitely. 
This  definition  applies  to  any  convex  cone  and  is  indepen- 
dent of  the  number  of  lateral  faces  of  the  original  pyramid. 

THE  LATERAL  AREA  OF  RIGHT  CIRCULAR  CONES 
160.  In  our  work  we  can  find  the  lateral  areas  of  right 
circular  cones  only.  Th.  97  gives  us  a  formula  for  the  lateral 
area  of  regular  pyramids  only,  and  not  for  all  pyramids. 
In  studying  the  lateral  areas  of  cones,  therefore,  we  will 
imagine  that  a  series  of  regular  pyramids  is  inscribed  in  a 
right  circular  cone,  and  that  the  number  of  sides  of  the 
pyramids  is  increased  indefinitely.  It  is  evident  that  a 
pyramid  will  soon  be  found  that  can  with  difficulty  be  dis- 
tinguished from  the  cone. 


130  SOLID   GEOMETRY 

The  following  theorem  from  higher  mathematics  will  be 
assumed  : 

THEOREM  115.  There  is  a  definite  limit  to  the  lateral 
areas  of  a  series  of  regular  pyramids  inscribed  in  a  right 
circular  cone  when  the  number  of  lateral  faces  is  increased 
indefinitely. 

The  lateral  area  of  a  right  circular  cone  is  defined  as  the 
limit  of  the  lateral  areas  of  a  series  of  regular  inscribed 
pyramids  as  the  number  of  lateral  faces  is  increased  indefi- 
nitely. Therefore,  since  the  lateral  area  of  a  regular  pyramid 
is  one-half  the  product  of  the  perimeter  of  the  base  and  the 
slant  height,  no  matter  how  many  lateral  faces  the  pyramid 
may  have,  we  will  assume 

THEOREM  116.  The  lateral  area  of  a  right  circular  cone 
is  one-half  the  product  of  the  perimeter  of  its  base  and 
its  slant  height. 

COR.  If  r  is  the  radius  of  the  base,  /  the  slant  height, 
L  the  lateral  area,  and  A  the  total  area  of  a  right  circular 
cone, 


161.  NOTE  1.     The   lateral   areas   of   cones    can  be  found    only 
approximately;  in  fact,  the  lateral  area  of  a  regular  inscribed  pyramid 
of  a  great  many  lateral  faces  is  taken  as  an  approximation  to  the  lateral 
area  of  a  right  circular  cone. 

162.  NOTE  2.     The  lateral  surface  of  a  right  circular  cone  may  be 
developed  into  a  sector  of  a  circle.     Cut  out  a  circle  whose  radius  is 
equal  to  the  slant  height  of  the  cone.     Wrap  the  circle  about  the  cone 
and  cut  it  so  that  the  sector  exactly  fits  about  the  cone  without  over- 
lapping.    Open  out  the  paper. 

If  r  is  the  radius  of  the  circle  and  a  the  angle  of  the  sector,  the  area 


of  the  sector  is  by  plane  geometry  -^-  or  -^77  •  M^!  that  is,  one-half 

ooU          obU 

the  product  of  the  radius  and  the  length  of  its  arc. 


AREAS  AND   VOLUMES  131 

THE   VOLUME   OF   CONES   WITH   CIRCULAR   BASES 

163.  In  studying  the  volumes  of  cones  we  will  imagine 
that  a  series  of  pyramids  is  inscribed  in  a  cone  with  a  cir- 
cular base,  and  that  the  number  of  lateral  faces  of  the  pyra- 
mid is  increased  indefinitely. 

The  following  theorem  from  higher  mathematics  will  be 
assumed  : 

THEOREM  117.  There  is  a  definite  limit  to  the  volumes 
of  a  series  of  pyramids  inscribed  in  a  cone  with  a  circular 
base  when  the  number  of  lateral  faces  of  the  pyramid  is 
increased  indefinitely. 

NOTE.  Theorem  117  is  true  no  matter  what  is  the  nature  of  the 
initial  pyramid  of  the  series. 

The  volume  of  the  cone  is  defined  as  the  limit  of  the 
volumes  of  a  series  of  inscribed  pyramids  as  the  number  of 
lateral  faces  is  increased  indefinitely.  ' 

Therefore,  since  the  volume  of  a  pyramid  is  one-third 
the  product  of  the  area  of  the  base  and  the  altitude;  no 
matter  how  many  lateral  faces  the  pyramid  may  have,  we 
will  assume 

THEOREM  118.  The  volume  of  a  cone  with  a  circular 
base  is  one-third  the  product  of  the  area  of  its  base  and  its 
altitude. 

NOTE.  As  proved  in  higher  mathematics,  Th.  118  is  true  for  any 
cone.  It  can  be  used  here,  however,  only  for  cones  with  circular  bases. 

COR.  If  r  is  the  radius  of  the  base  of  a  cone,  h  its  alti- 
tude, and  V  its  volume, 


164.  NOTE  1.  The  volume  of  cones  can  be  found  only  approxi- 
mately; in  fact,  the  volume  of  an  inscribed  pyramid  of  a  great  many 
lateral  faces  is  taken  as  an  approximation  to  the  volume  of  the  cone. 

166.  NOTE  2.  By  similar  processes  the  lateral  area  and  the  volume 
of  cones  can  be  obtained  from  circumscribed  pyramids. 


132  SOLID   GEOMETRY 

THE   MEASUREMENT   OF  FRUSTUMS 
OF   CONES 

166.  The  theorems  for  the  measurement  of  the  lateral 
surface  and  the  volume  of  certain  frustums  of  cones  may  be 
obtained  from  the  corresponding  theorems  concerning  the 
lateral  surface  and  the  volume  of  frustums  of  pyramids. 
The  method  is  similar  to  that  used  in  obtaining  the  theorems 
for  the  lateral  surface  and  the  volume  of  certain  cones  from 
the  corresponding  theorems  concerning  pyramids. 

If  the  bases  of  a  frustum 
of  a  pyramid  are  inscribed 
in  the  bases  of  a  frustum  of 
a  cone,  and  the  lateral  edges 
of  the  frustum  of  the  pyr- 
amid -are  elements  of  the 

frustum    of   the  cone,   the 

frustum  of  the  pyramid  is  FlG-  151 

said  to  be  inscribed  in  the  frustum  of  the  cone. 

By  a  process  similar  to  that  used  on  pages  125  and  126, 
the  number  of  lateral  faces  of  the  inscribed  frustum  may  be 
increased  indefinitely.  It  is  evident  that  a  frustum  of  a 
pyramid  will  soon  be  obtained  that  can  with  difficulty  be 
distinguished  from  the  frustum  of  the  cone. 

167.  The  following  theorems  will  be   assumed   without 

further  discussion: 

• 

THEOREM  119.  If  rl  and  r2  are  the  radii  of  the  upper  and 
lower  bases  of  a  frustum  of  a  right  circular  cone,  and  /  is 
its  slant  height,  the  lateral  area  of  the  frustum  is 

7r/(r1+r2) 

THEOREM  120.  If  r1  and  r2  are  the  radii  of  the  upper 
and  lower  bases  of  a  frustum  of  a  cone  with  a  circular  base, 
and  h  is  its  altitude,  the  volume  of  the  frustum  is 


7rr22)  =1/^/i7r(r]  2+r22+r,r,) 


AREAS  AND  VOLUMES  133 

EXERCISES    INVOLVING    THE    AREAS    AND    VOLUMES    OF 
CYLINDERS,   CONES,   AND   FRUSTUMS 

168.  1.  Find  the  total  area  and  the  volume  of  a  right  circular 
cylinder  30  ft.  high  and  8  ft.  in  diameter. 

2.  Find  the  cost  of  digging  a  cistern  25  ft.  deep  and  6  ft.  in 
diameter  at  5^  a  cubic  foot. 

3.  How  many  gallons  of  paint  are  required  for  the  lateral  sur- 
face of  a  cylindrical  tower  100  ft.  high  and  30  ft.  in  diameter? 
Allow  1  gal.  of  paint  for  500  sq.  ft.  of  surface. 

4.  Find  the  total  area  of  a  right  circular  cone  whose  slant 
height  is  15  in.  and  whose  radius  is  6  in. 

5.  Find  the  total  area  and  volume  of  a  right  circular  cone  if 
its  altitude  is  12  in.  and  its  diameter  is  18  in. 

6.  Find  the  total  area  and  volume  of  a  right  circular  cone  if 
its  slant  height  is  13  in.  and  its  radius  5  in. 

7.  What  must  be  the  depth  of  a  cylindrical  measure  12  in.  in 
diameter  if  it  holds  1  bushel?  ,  2150.42  in.  =  1  bu. 

8.  A  cylindrical  tower  has  a  conical  top.     Find  the  total  cost 
of  painting  the  tower 'at  25  ^  a  square  foot.     The  total  height  of 
the  tower  is  50  ft.,  the  height  of  the  cylindrical  part  is  42  ft.,  and 
the  diameter  is  18  ft. 

9.  Find  the  total  area  and  the  volume  of  a  frustum  of  a  right 
circular  cone  if  the  radii  of  its  bases  are  3  in.  and  10  in.  respectively 
and  its  slant  height  is  5  in. 

10.  Find  the  total  area  and  the  volume  of  the  solid  formed  by 
revolving  a  square  about  one  side  if  one  side  is  (1)  15  in.;  (2)  5. 

11.  Find  the  total  area  and  the  volume  of  the  solid  formed  by 
revolving  a  rectangle  whose  sides  are  5  in.  and  7  in.  (1)  about  the 
side  5  in.;   (2)  about  the  side  7  in. 

12.  What  are  the  answers  called  for  in  Ex.  11  if  the  sides  of  the 
rectangle  are  a  and  b? 

13.  What  are  the  answers  called  for  in  Exs.  11  and  12  if  a 
right  triangle  whose  legs  are  7  in.  and  5  in.  is  used  instead  of  a 
rectangle? 

14.  Find  the  total  area  and  the  volume  of  the  solid  formed  by 
revolving  an  isosceles  triangle  about  its  base,  if  the  base  is  8  in. 
and  one  leg  is  10  in.  . 


134 


SOLID   GEOMETRY 


15.  Answer  Ex.  14  if  the  triangle  is  revolved  about  its  altitude. 

16.  Find  the  total  area  and  the  volume  of  the  solid  formed  by 
the  revolution  of  an  isosceles  right  triangle  about  the  hypotenuse 
(1)  if  one  leg  is  6  in.;   (2)  if  one  leg  is  a. 

17.  ABC  is  an  isosceles  right  triangle  each  of  whose  legs  is  3  in. 
/  is  a  line  perpendicular  to  hypotenuse  AB  at  B.    Find  the  total 
area  and  the  volume  of  the  solid  formed  by  revolving  A  ABC 
about  /. 

GENERAL  FORMULA 
LATERAL  AREA  OF  SOLIDS   OF  REVOLUTION 

169.  THEOREM  121.  The  area  of  the  surface  generated 
by  a  segment  revolving  about  an  axis  in  The  same  plane 
with  it,  but  not  crossing  it,  is  the  product  of  the  projection 
of  the  line  segment  on  the  axis  and  the  circumference  of 
a  circle  whose  radius  is  the  perpendicular  erected  at  the 
mid-point  of  the  segment  and  terminated  by  the  axis. 


D 


No.  1 


No.  2 
FIG.  152 


No.  3 


Hypothesis:  AB  is  a  line  segment  revolving  about  an 
axis  which  is  in  the  same  plane  with  AB,  but  does  not  cross 
AB.  CD  is  the  projection  of  AB  on  the  axis.  E  is  the 
mid-point  of  AB.  EF  _L  AB  and  is  terminated  by  CD. 
Let  L  represent  the  area  generated  by  AB. 

Conclusion:    L  =  CD  •  2irEF. 

Case  A.     When  AB  is  parallel  to  the  axis,  L  is  the  lateral 
area  of  a  right  circular  cylinder  (Fig.  152,  No.  1). 
The  proof  is  left  to  the  pupil. 


AREAS  AND  VOLUMES  135 

Case  B.  When  A B  meets  the  axis  at  A,  L  is  the  lateral 
area  of  a  right  circular  cone  (Fig.  152,  No.  2). 

For  right  circular  cones  L  =  ABirBD,  or  AB2irEG. 

To  prove  L  =  CD-  2irEF,  prove  CD  -  2irEF  =  AB  •  2irEG. 

That  is,  AB  •  EG  must  be  replaced  by  CD  •  EF. 

Use  similar  triangles  ABD  and  EFG. 

Let  the  pupil  complete  the  proof. 

Case  C.  When  AB  is  not  parallel  to  the  axis  and  does 
not  meet  the  axis,  L  is  the  lateral  area  of  a  frustum  of  a 
right  circular  cone  (Fig.  152,  No.  3) . 

The  lateral  area  of  a  frustum  of  a  right  circular  cone  is 

ir/(rH-rt). 

Since  EG  =  %(r\-\-rz),  L  =  AB  •  2irEG. 
To  prove  L  =  CD-  2wEF,  prove  CD  •  2wEF  =  AB  •  27r£C. 
That  is,  replace  AB  •  EG  by  CD  •  £F. 

Let  the  pupil  complete  the  proof.     Draw  AH  from  A  JL  BD. 

Ex.  1.  Fig.  153  shows  half  of  a  regular  hexagon 
inscribed  in  a  semicircle.  Find  the  area  of  the 
surface  generated  by  the  broken  line  A  BCD  revolv- 
ing about  the  diameter  AD.  The  radius  of  the 
circle  is  3  in. 

Ex.  2.  What  would  be  the  area  of  the  surface 
formed  in  Ex.  1  if  A  BCD  were  circumscribed  about 
the  semicircle? 

170.  NOTE.     The  lateral  area  of  a  frustum  of  a         FlG'  153 
right  circular  cone  may  be  found  from  its  development.     Show  that 
this  development  is  a  sector  of  a  circular  ring  (Fig.  154). 

For  the  area  of  the  sector  of  the  circular 
ring  we  have  (a  is  the  angle  of  the  sector) : 

Area  =^  (TTrJ-irr& 


.  2ira 
ri)  360  Cr2+ri)     «"»     "'  L  360    T  360"  J 


136 


SOLID   GEOMETRY 


GENERAL   FORMULA  FOR   THE   VOLUMES   OF   SOME 
SOLIDS    OF   REVOLUTION 

171.  THEOREM  122.  The  volume  of  a  solid  generated 
by  a  triangle  that  revolves  about  an  axis  in  its  plane,  passing 
through  one  of  its  vertices,  but  not  crossing  it,  is  one-third 
the  product  of  the  area  of  the  surface  generated  by  the  side 
opposite  the  fixed  vertex  and  the  corresponding  altitude  of 
the  triangle. 

Case  A.  When  one  side  of  the  triangle  coincides  with 
the  axis. 


No.  1 


Hypothesis:  A  ABC  reyolves  about  side  AB  as  an  axis. 
The  axis  is  supposed  to  pass  through  A.  BC  is  the  side 
opposite  A.  h  is  the  altitude  from  A  to  BC.  r  is  the  _L  from 
C  to  AB.  V  is  the  volume  generated  by  A  ABC. 

Conclusion:     V  =  Hh  •  area  generated  by  BC. 

Analysis  and  construction  (Fig.  155,  Nos.  1  and  2): 
I.  a.  The  volume  generated  by    A  ABC  is 

(1)  y3TrrzBX+l/3irr2AX,  or  Xirr*AB. 

(2)  Kvr*BX-Kvr*AXtorXirr*AB. 
b.  The  area  generated  by  BC  is  irrBC. 
.'.  we  are  to  prove  V  =  Yzk  •  wrBC. 

II.    /.   prove  Hh-  irrBC  =  lA irr2AB. 

III.  /.  prove  h-BC  =  r-  AB. 

AB    h 

IV.  .'.  prove  g£  =  -. 

V.    /.  prove  AABD  ~  ABCX. 


AREAS  AND   VOLUMES  137 

Ex.  1.  Give  the  analysis  and  proof  for  the  case  in  which  B 
is  taken  as  the  fixed  point  instead  of  point  A . 

Ex.  2.  Apply  the  theorem  to  the  special  case  (Fig.  155)  in 
which  BC  is  perpendicular  to  the  axis. 

Case  B.  When  no  side  of  the  triangle  coincides  with  the 
axis. 

Hypothesis:  AABC  revolves  about  line  /  as  an  axis. 
Line  /  is  in  plane  ABC,  passes  through  vertex  A,  but  does 
not  cross  the  triangle.  BC  is  the  side  of  A  ABC  opposite 
vertex  A.  h  is  the  altitude  from  A  to  BC.  V  represents 
the  volume  generated  by  AABC  (Fig.  155,  No.  3). 

Conclusion:    -V  =  Hh  •  area  generated  by  BC. 

Analysis  and  construction:  To  find  the  volume  generated 
by  A  ABC,  extend  CB  to  meet  the  axis  at  E  and  subtract  the 
volume  generated  by  &EBA  from  the  volume  generated  by 
AECA. 

Outline  of  proof: 

I.  Volume  generated  by  AECA  =  Hh  •  area  CE. 
II.  Volume  generated  by  AEBA=%h  •  area  BE. 
.'.  volume  generated  by  ABCA=Hh  •  area  CB. 


FIG.  156 


Ex.  3.  Apply  the  theorem  to  the  special  cases  (Fig.  155,  No. 
3)  in  which  (1)  BC  is  parallel  to  /;  (2)  BC  extended  cuts  /  below  A . 

Ex.  4.  Describe  the  figure  formed  in  each  case  of  Th.  122 
and  in  Exs.  1,2,  and  3. 


10 


138  SOLID   GEOMETRY 

VOLUMES  BY   CAVALIERTS  THEOREM 

172.  NOTE.    The  treatment  of  volumes  that  follows  may  be  sub- 
stituted for  the  treatment  in  §§142,  144,  156,  and  163  if  the  teacher 
desires. 

CAVALIERTS   THEOREM 

173.  The  following  theorem  is  proved  in  higher  mathe- 
matics.    It  will  be  assumed  here. 

THEOREM  123.  If  two  solids  lie  between  parallel  planes, 
and  if  the  two  sections  made  by  any  plane  parallel  to  then- 
bases  are  equivalent,  then  the  solids  are  equivalent. 

VOLUME   OF  ANY  PRISM 

174.  THEOREM  104.    The  volume  of  any  prism  is  the 
product  of  the  area  of  its  base  and  its  altitude. 


FIG.  157 

Hypothesis:  P  is  any^prism  with  base  b,  altitude  ht  and 
V  its  volume. 

Conclusion:     V  of  P  =  bh. 

Analysis  in  general:  To  find  the  V  of  P,  compare  P 
with  a .  rectangular  parallelepiped  that  has  an  equivalent 
base  and  the  same  altitude. 

Construction:  Construct  P'  a  rectangular  parallelepiped 
with  base  b'  =  6,  and  an  edge  h'  =  h.  Let  P'  and  P  stand  on 
the  same  plane.  Pass  a  plane  parallel  to  6  and  b'  cutting 


both  P'  and  P. 
respectively. 


Let  sf  and  5  be  the  sections  of  P'  and  P 


AREAS  AND   VOLUMES 


139 


Analysis: 

I.  To  prove  V  of  P  =  bh,  prove  P'  =  P. 
II.  To  prove  P'  =  P,  prove  the  sections  of  P  and  P' 
made  by  planes  parallel  to  the  bases  are  equivalent. 
III.   /.  prove  5  =  5'. 

Outline  of  proof: 
I.  s'  =  b'  =  b  =  s.     :.  s'  =  s. 

II.  In  the  same  way,  the  sections  of  P  and  P'  made  by 
every  plane  parallel  to  the  bases  are  equivalent. 

III.  /.  P'=P. 

IV.  VotP'  =  b'h'. 

V.   .'.  V  oiP  =  b'ti  =  bh. 
NOTE.     If  this  proof  of  Th.  104  is  used,  §§136-141  may  be  omitted. 

EQUIVALENT  TRIANGULAR  PYRAMIDS 

176.  THEOREM  106.      If  two  triangular  pyramids  have 
equivalent  bases  and  equal  altitudes,  they  are  equivalent. 


FIG.  158 

Hypothesis:    P  and  P'  are  two  pyramids  with  equivalent 
bases  b  and  b'  and  equal  altitudes  h  and  h'. 
Conclusion :    P  =  P'. 

Analysis  and  construction:  Suppose  P  and  P'  are  stand- 
ing on  the  same  plane.  Pass  any  plane  through  P  and  P' 
parallel  to  b  and  b',  cutting  P  in  section  c,  and  P'  in  section 
c',  and  prove  c  =  cf. 

NOTE.  At  this  point  the  theorems  concerning  the  volumes  of 
pyramids  follow  as  given  in  §§145-147. 


140 


SOLID   GEOMETRY 


THE   VOLUME   OF   CYLINDERS  WITH   CIRCULAR  BASES 

176.  We  know  that  it  is  not  possible  to  construct  with 
straight   edge   and   compass,    only,    a   triangle,   square,   or 
rectangle  that  has  the  same  area  as  a  given  circle.     We  do 
know,  however,  that  there  are  instruments  by  which  these 
figures  can  be  constructed.     This  must  be  assumed  in  the 
next  two  proofs. 

177.  THEOREM  114.     The  volume  of  a  cylinder  with  a 
circular  base  is  the  product  of  the  area  of  its  base  and  its 
altitude. 


FIG.  159 

Hypothesis:  C  is  a  cylinder  with  a  circular  base  b;  h  is 
its  altitude;  V  is  its  volume. 

Conclusion:     V  of  C  =  bh. 

Analysis  in  general:  To  find  V  of  C,  compare  it  with  a 
prism  that  has  an  equivalent  base  and  the  same  altitude. 

Let  the  pupil  give  the  construction,  complete  the  analysis,  and  give 
proof.  Use  §174  as  a  model. 

THE   VOLUME    OF   CONES   WITH   CIRCULAR  BASES 

178.  THEOREM  118.  The  volume  of  a  cone  with  a  cir- 
cular base  is  one-third  the  product  of  the  area  of  its  base 
and  its  altitude. 

Analysis:  To  find  the  volume  of  the  cone,  compare  it 
with  a  pyramid  that  has  an  equivalent  base  and  the  same 
altitude. 


AREAS  AND   VOLUMES  141 

MISCELLANEOUS   EXERCISES 

179.  1.  A  sheet  of  paper  6  in.  X8  in.  is  bent  into  a  right  circular 
cylinder.    Find  the  total  area  and  the  volume.    (Two  answers.) 

2.  Water  is  flowing  through  a  pipe  2  in.  in  diameter  at  the 
rate  of  150  ft.  per  minute.     How  many  cubic  feet  is  that  per  hour? 

3.  A  cylindrical  tomato  can  is  4T%  in.  high  and  4  in.  in  diam- 
eter.    Find  its  capacity  in  quarts.     231  cu.  in.  =  1  gal. 

4.  A  regular  four-sided  prism  is  inscribed  in  a  right  circular 
cylinder.     Find  the  volume  and  the  lateral  area  of  the  prism  if 
the  radius  of  the  cylinder  is  10  in.  and  its  altitude  is  25  in. 

5.  The  section  of  a  right  circular  cone  made  by  a  plane  through 
the  altitude  is  a  triangle  whose  base  angles  are  45°.     If  the  height 
of  the  cone  is  6  in.,  find  the  total  area  and  the  volume. 

6.  The  volume  of  an  irregular  body  that  will  not  absorb  water 
may  be  found  by  placing  it  in  water  and  finding  the  volume  of  the 
water  displaced.     What  is  the  volume  of  a  stone  if,  when  it  is 
dropped  into  a  cylindrical  tank  2  ft.  in  diameter,  it  causes  the 
water  in  the  tank  to  rise  2  in.? 

7.  A  regular  hexagonal  pyramid  is  inscribed  in  a  right  circular 
cone  whose  radius  is  10  in.  and  whose  height  is  24  in.     Find  the  total 
area  and  the  volume  of  the  pyramid. 

8.  Find    the    lateral   area    of   the 
frustum  of   a  cone  formed  by  rolling 
the  sector  of  the  circle  shown  in  Fig. 
160.     Use  the  data  given. 

9.  The   altitude  of  a  cone    equals 

the  diameter  of  its  base.  Using  2r  for  FlG-  16° 

the  diameter,  find  formulas  for  the  volume  and  for  the  total  area. 

10.  Find  a  formula  for  the  volume  of  a  hollow  column  in  the 
form   of   a   right   circular  cylinder.      Use  h  =  height, 

t  =  thickness,  and  d  =  outside  diameter.     The  column     /*    "\ 
is  open  at  both  ends. 

11.  The  cross  section  of  a  straight  tunnel  %  mi. 
long  is  of  the  form  shown  in   Fig.    161.     Find   the 
quantity   of   material   taken   out,   using   the   dimen- 


sions given  in  the  diagram.      Height  of  rectangular       20'~ 
part,  30  ft.  FIG.  161 


142  SOLID   GEOMETRY 

SPHERICAL   MEASUREMENTS 
AREAS   OF   SPHERES 

180.  We  have  seen  in  plane  geometry  that  if  a  regular 
polygon  is  inscribed  in  a  circle,  and  if  the  number  of  sides 
is  increased  indefinitely,  a  polygon  is  soon 
formed  which  can  with  difficulty  be  distin- 
guished from  the  circle.  (See  Plane  Geometry, 
§295.)  The  perimeter  of  such  a  polygon  may 
be  taken  as  an  approximation  to  the  circum- 
ference of  the  circle.  An  approximation  to  the 
surface  of  the  sphere  may  be  obtained  in  a 
somewhat  similar  way. 

Suppose  a  semicircle  is  divided  into  any 

T^rr*     1  fi9 

number  of  equal  arcs  and  the  points  of  division 
are  joined.  A  chain  of  equal  chords  is  obtained  which  is  half 
of  an  inscribed  regular  polygon,  with  two  of  its  vertices  at 
the  ends  of  the  diameter  XY  (Fig.  162).  If  we  bisect  the 
original  arcs  and  join  the  points  of  division,  half  of  a  regular 
polygon  of  twice  as  many  sides  will  be  inscribed  in  the  semi- 
circle. This  process  can  be  continued  indefinitely.  If  now 
we  revolve  the  figure  about  the  diameter  XY  as  an  axis,  we 
know  that  the  semicircle  will  generate  a  spherical  surface. 
The  chain  of  equal  chords  will  generate  the  lateral  surface  of 
a  series  of  cones  and  frustums  of  cones  inscribed  in  the  sphere. 
We  have  seen  that  if  the  semicircle  is  divided  into  a  very 
great  number  of  equal  arcs,  the  chain  of  equal  chords  obtained 
can  with  difficulty  be  distinguished  from  the  semicircle.  It 
is  evident,  therefore,  that  if  this  semicircle  is  revolved  about 
its  diameter,  the  lateral  surface  of  the  sets  of  cones  and  frus- 
tums obtained  can  with  difficulty  be  distinguished  from  the 
spherical  surface. 

The  measure  of  the  surface  of  one  of  these  sets  of  inscribed 
solids  may  be  taken  as  an  approximation  to  the  measure  of 
the  surface  of  the  sphere. 


AREAS  AND   VOLUMES 


143 


181.  PROBLEM.  To  find  the  area  of  the  surface  gener- 
ated by  the  revolution  of  a  chain  of  equal  chords  inscribed 
in  a  semicircle  as  it  revolves  about  the  diameter  of  that 
semicircle  as  an  axis. 


FIG.  163 

Given  AB+BC+CD+  etc.,  a  chain  of  equal  chords 
inscribed  in  the  semicircle  ADG,  revolving  about  the  diam- 
eter AG  as  an  axis. 

To  find  the  area  of  the  surface  generated  by  AB -\-BC-\- 
CD+  etc. 

Analysis  in  general: 

I.  To  find  the  area  of  the  surface  generated  by  AB  + 
BC+CD+etc.,  find  the  area  of  the  surface  gen- 
erated by  AB,  by  BC,  by  etc.,  and  add  the  results. 
II.  To  find  the  area  of  surfaces  generated  by  AB,  BC, 

CD,  +  etc.,  use  Th.  121. 

Construction:     .'.    draw  the   J_s  from  B,  C,  D,  etc.,  to 
the  diameter  AG,  meeting  the  diameter  in  X,  Y,  0,  etc., 
respectively.     Draw  the  _L  bisectors  of  chords  AB,  BC,  etc. 
Outline  of  solution: 


XY. 
....  etc. 
HO  =  KO  =  etc. 

/.  area  (AB+BC '+etc.)  =2wHO(AX+XY+etc.) 
or  area  (AB+BC+etc.)  =2irHO  •  AG. 


144  SOLID   GEOMETRY 

182.  The   following   theorem   from   higher  mathematics 
will  be  assumed: 

THEOREM  124.  The  areas  of  the  surfaces  generated  by  a 
series  of  chains  of  equal  chords  inscribed  in  the  same  semi- 
circle, and  revolving  about  the  diameter  of  that  semicircle 
as  an  axis,  have  a  definite  limit  if  the  number  of  chords 
is  increased  indefinitely. 

183.  Since  we  cannot  measure  any  spherical  surface  in 
terms  of  plane  units,  we  shall  have  to  define  what  is  meant 
by  the  area  of  a  spherical  surface. 

The  area  of  a  spherical  surface  is  defined  as  the  limit 
of  the  areas  of  the  surfaces  generated  by  a  series  of  chains 
of  equal  chords  inscribed  in  the  same  semi- 
circle, and  revolving  about  the  diameter  of 
that  semicircle  as  an  axis  as  the  number  of 
chords  is  increased  indefinitely. 

184.  Therefore,  since  the  area  of  the  sur- 
face generated  by  a  chain  of  equal  chords 
inscribed    in    a    semicircle,    and   revolving 
about  the  diameter  of  that  semicircle  as  an 
axis,  is  always  2wHO  -  XY  (Fig.  164),  no 
matter  how  many  chords  there  may  be  in 

the  chain,  we  will  assume  FlG-  164 

THEOREM  125.  The  area  of  a  spherical  surface  is 
2?rr  •  d  when  r  represents  the  radius  of  the  sphere,  and  d 
its  diameter. 

Since  2  TIT  represents  the  circumference  of  a  great  circle 
of  the  sphere,  Th.  125  may  be  stated: 

The  area  of  a  spherical  surface  is  the  product  of  its 
diameter  and  the  circumference  of  a  great  circle. 

If  we  substitute  2r  for  d  in  the  first  statement  in  Th.  125, 
we  have : 

The  area  of  the  spherical  surface  =4?rr2. 


AREAS  AND  VOLUMES  145 

Since  ?rr2  represents  the  area  of  a  great  circle,  Th.  125 
may  be  stated: 

The  area  of  a  sperical  surface  is  equal  to  the  area  of 
four  great  circles.  . 

Ex.  1.  Find  the  area  of  a  sphere  whose  radius  is  3  in.;  4}/2  in. 

Ex.  2.  If  the  area  of  a  sphere  is  265  sq.  in.,  find  its  radius. 

Ex.  3.  Find  a  formula  for  obtaining  the  radius  of  a  sphere  from 
its  area. 

Ex.  4.  A  tank  consists  of  a  cylindrical  portion  with  hemispher- 
ical ends.  The  diameter  of  the  ends  equals  the.  diameter  of  the 
cylinder,  which  is  2  ft.  The  total  length  is  7  ft.  Find  the  total 
area. 

AREAS    OF   ZONES 

185.  The  portion  of  a  spherical  surface  included  between 
two  parallel  planes  is  called  a  zone  (Fig.  165). 

The    circles   in   which    the    parallel  A 

planes  cut  the  sphere  are  called  the 
bases  of  the  zone.  The  perpendicular 
distance  between  the  parallel  planes  is 
called  the  altitude  of  the  zone. 

If  one  of  the  planes  is  tangent  to  the 
sphere,  the  zone  is  called  a  zone  of  one 
base. 

Just  as  a  spherical  surface  is  considered  as  generated  by 
the  revolution  of  a  semicircle  about  a  diameter  as  an  axis, 
so  a  zone  is  considered  as  generated  by  the  revolution  of 
an  arc  of  that  semicircle  about  the  same  diameter  as  an  axis. 

In  Fig.  165  the  revolution  of  the  semicircle  ABCD  about 
the  diameter  AD  generates  the  spherical'  surface.  The 
revolution  of  AB  about  AD  generates  a  zone  of  one  base; 
the  revolution  of  BC  about  AD  generates  a  zone  of  two 
bases  BCEF.  XY  is  the  altitude  of  zone  BCEF.  The 
circles  BF  and  CE  are  the  bases  of  the  zone  BCEF. 

Exercise.  What  are  the  bases  of  the  North  Temperate  Zone 
of  the  earth?  Of  the  Arctic  Zone? 


146 


SOLID   GEOMETRY 


FIG.  166 


186.  COR.    The  area  of  a  zone  is  2irr  times  the  altitude 
of  the  zone. 

The  corollary  may  be  verified  by  the 
same  method  as  was  used  in  verifying  Th. 
125.  The  arc  AD  (Fig.  166)  is  divided  into 
any  number  of  equal  parts,  and  the  points 
of  division  joined.  Find  the  area  generated 
by  AB+BC+  etc.  Imagine  the  number  of 
divisions  to  be  increased  indefinitely. 

Since  2irr  is  the  circumference  of  a  great 
circle,  the  corollary  may  read : 

The  area  of  a  zone  is  the  product  of  the  altitude  of  the 
zone  and  the  circumference  of  a  great  circle. 

If  h  is  the  altitude  of  the  zone  and  L  its 
area,  we  have  L  =  2irrh 

Ex.  1.  The  radius  of  a  sphere  is  10  in.  and 
the  altitude  of  a  zone  is  6.  Find  the  area  of 
the  sphere  and  of  the  zone. 

Ex.  2.  The  diameter  of  a  circle  (Fig.  167)  is 
6  in.  AB  subtends  a  central  angle  of  60°.  Find 
the  area  of  the  zone  of  one  base  formed  if  AB 
revolves  about  AD. 

Ex.  3.  In  Fig.  168,  the  diameter  of  circle  0  is 
51".  The  chord  BD  is  24".  Find  the  area  of  the 
zone  generated  if  AB  revolves  about  AC. 

Suggestion.  Draw  BC  and  use  Th.  108,  Plane 
Geometry. 

AREAS   OF  LUNES 

187.  We  have  defined  a  lune  as  a  spherical 
polygon  of  two  sides.     In  other  words,  it  is 
a  portion  of  a  sphere  between  two  semi-great 
circles  (Fig.  169). 

The  following  corollary  of  this  definition 
is  evident:  FIG.  169 

COR.    The  two  angles  of  a  lune  are  equal. 


AREAS  AND  VOLUMES  147 

188.  Since  the  stun  of  the  possible  adjacent  spherical 
angles  that  have  a  common  vertex  on  the  sphere  is  360°, 
we  will  assume  that  the  surface  of  the  sphere  can  be  divided 
into  360  equal  parts  by  great  circles  that  have  a  common 
diameter.     Any  two  of  these  semi-great  circles  form  a  lune. 
We  will  assume  that  the  ratio  of  the  surface  of  this  lune  to 
the  surface  of  the  sphere  is  equal  to  the  ratio  of  the  number 
of  degrees  in  the  angle  of  the  lune  to  360°.     If,  however,  a 
lune  is  drawn  at  random  on  the  sphere,  its  angle  may  be 
incommensurable  with  360°.      It  can  be  proved  that  the 
same  relation  holds  in  this  case.     We  have,  therefore, 

As.  30.    The  area  of  a  lune  is  to  the  area  of  a  sphere  as 
the  angle  of  the  lune  is  to  360°. 

If  L  represents  the  area  of  a  lune,  a  the  measure  of  its 

angle,  and  A  the  area  of  the  sphere,  we  have  -r 

Solving  for  L, 

T  _  *    a         a      A     2  _  a7rf2 
360"  360*  ="90" 

189.  A  birectangular  spherical  triangle 
whose  vertex  angle  is  one  degree  is  called 
a  spherical  degree  (Fig.  170). 

It  is  evident  that  a  spherical  degree  is  FlG-  17° 

half  a  lune  whose  angle  is  one  degree.  A  spherical  degree 
is,  therefore,  ^20  of  the  surface  of  the  sphere.  Its  area  is 
constant  for  any  given  sphere. 

The  following  theorem  is,  therefore,  evident: 

THEOREM  126.    The  area  of  a  lune  in  spherical  degrees 
is  equal  to  twice  the  measure  of  its  angle  in  angle  degrees. 

If  L  represents  the  area  of  the  lune,  and  a  the  measure  of 
its  angle,  L  =  2a  spherical  degrees 

But  a  spherical  degree  is  =^:  .  47rr2. 


2a  a  in-2 


148  SOLID   GEOMETRY 

AREAS    OF   SPHERICAL   TRIANGLES 

190.  THEOREM  127.  The  area  of  a  spherical  triangle 
expressed  in  spherical  degrees  is  equal  to  the  spherical 
excess  of  the  triangle. 


Hypothesis:    ABC  is  a  spherical  triangle.     A,  B,  and  C 
are  the  measures  of    /.A,    /.B,  and    Z.C  respectively. 

Conclusion:     The   area   of    AABC  is    (A+B+C-18Q) 
spherical  degrees. 

Analysis  and  construction: 

I.  To  find  the  area  of  A  ABC,  compare  A  ABC  with 

the  lunes  whose  angles  are  /.A,   /.B,  and  Z.C. 

II.  /.  complete  the  great  circles  ABA'B',  ACA'C',  and 

BCB'C'. 

Outline  of  proof: 

I.  a.         Area  of  lune  CAC'B  =  2C  sph.  deg. 

b.  Area  of  lune  BCB'A  =  2B  sph.  deg. 

c.  Area  of  lune  ACAfB  =  2A  sph.  deg. 

II.  a.  Or  area  Al+area  AIV  =  2C  sph.  deg. 

b.  Area  Al+area  All  =  25  sph.  deg. 

c.  Area  Al+area  ACBA'  =  2A  sph.  deg. 

III.  a.  ACBA'and  AIII  are  symmetric. 

b.  .'.  ACBA'=AIII. 

c.  .'.  area  Al+areaAHI  =  2A  sph.  deg. 

IV.  Adding, 

Areas  (3AI  + AII  + AIII  + AIV)  =2(,4+B+Q  sph. deg. 


AREAS  AND  VOLUMES  149 

V.   AI  +  AII  +  AIII  +  AIV  make  one  hemisphere. 

/.  area  (AI  +  AII  +  AIII  +  AIV)  =360  sph.  deg. 
VI.    /.  2  area  Al+360  sph.  deg.  =  2(A+B+Q  sph.  deg. 

/.  area  Al  +  180  sph.  deg.  =  (A+B+C)  sph.  deg. 
VII.    /.  area  AI  =  G4+5+C-180)  sph.  deg. 

COR.    If  r  represents  the  area  of  the   sphere,  £  the 
spherical  excess  of  the  triangle,  and  A  its  area, 


EXERCISES   INVOLVING   AREAS   OF   LUNES   AND 
SPHERICAL   TRIANGLES 

191.  1.  Find  the  area  of  a  lime  on  a  sphere  if  the  radius  of  the 
sphere  is  6  in.  and  the  angle  of  the  lime  is  25°;  50°;   60°;   120°. 

2.  Find  the  area  of  a  spherical  triangle  in  spherical  degrees  if 
the  angles  of  the  triangle  are  : 

a.  GO0,  150°,  120°.  c.   140°,  72°,  60°. 

b.  120°,  75°,  150°.  .   </.  72°,  65°,  90°. 

3.  Find  the  area  of  a  spherical  triangle  on  a  sphere  whose 
radius  is  18  in.  if  the  angles  of  the  triangle  are: 

a.  60°,  72°,  120°.  c.  45°,  90°,  105°. 

b.  50°,  80°,  150°.  rf.  68°,  74°,  96°. 

4.  The  area  of  a  lune  is  154  sq.  in.     Find  its  angle  if  the  radius 
of  the  sphere  is  8  in. 

5.  The  area  of  a  lune  whose  angle  is  75°  is  65  sq.  in.     Find  the 
area  and  the  radius  of  the  sphere. 

6.  What  is  the  angle  of  a  lune  whose  area  is  H',    H\   %\   % 
the  area  of  the  sphere? 

7.  The  area  of  a  trirectanguJar  spherical  triangle  is  ^  the 
surface  of  the  sphere. 

8.  How  many  degrees  in  the  angle  of  a  lune  if  its  area  is  equal 
to  the  area  of  a  great  circle  of  the  sphere? 

9.  What  portion  of  a  sphere  is  covered  by  a  spherical  triangle 
whose  angles  are  68°,  72°,  and  124°? 

10.  What  is  the  area  in  spherical  degrees  of  a.  spherical  tri- 
angle if  the  sides  of  its  polar  are  30°,  75°,  54°? 


150 


SOLID   GEOMETRY 


THE  VOLUME   OF   THE   SPHERE 

192.  If  a  chain  of  a  great  many  equal  chords  be  inscribed 
in  a  semicircle,  and  this  chain  of  chords  be  revolved  about 
the  diameter  of  this  semicircle  as  an  axis,  the  surface  gen- 
erated by  the  chain  of  chords  will  inclose  a  solid  whose 
volume  may  be  taken  as  an  approximation  to  the  volume 
of  the  sphere. 

PROBLEM.  To  find  the  volume  of  the  solid  inclosed  by 
the  surface  generated  when  a  chain  of  equal  chords  inscribed 
in  a  semicircle  revolves  about  the  diameter  of  that  semi- 
circle as  an  axis. 


FIG.  172 

Analysis  and  construction: 

I.  To  find  the  volume  inclosed  by  the  surface  generated 
by  AB+BC+CD+etc.,  join  the  vertices  A,  B, 
C,  D,  etc.,  with  0,  and  find  the  volume  generated 
by  AAOB,  ABOC,  ACO£>,  etc.    Add  the  results. 
II.   .'.  draw  the  J_s  from  0  to  AB,  EC,  etc.    (Th.  122). 
Outline  of  solution: 
I.  a.  Vol.  A05  =  area  AB  •  ysOX. 

b.  Vol.  50C  =  area  BC-  HOY. 

c.  Vol.  COD  =area  CD  •  HOZ. 
....  etc. 

II.  a.  Adding,  since  OX  =  OY  =  OZ, 

b.  Vol.  (AOB  +  BOC  +  COD  +  etc.)  = 
\OX  -  area(AB+ BC+CD+etc,). 


AREAS  AND   VOLUMES  151 

The  solution  obtained  to  the  foregoing  problem  may  be 
translated  thus: 

THEOREM  128.  The  volume  inclosed  by  the  surface 
generated  when  a  chain  of  equal  chords  inscribed  hi  a 
semicircle  revolves  about  the  diameter  of  that  semicircle 
as  an  axis  is  the  product  of  the  area  of  that  surface  and  one- 
third  the  radius  of  the  semicircle. 

193.  The  following  theorem   from  higher  mathematics 
will  be  assumed: 

THEOREM  129.  The  volumes  of  the  solids  inclosed  by 
the  surfaces  generated  when  a  series  of  chains  of  equal 
chords  inscribed  in  a  semicircle  revolves  about  the  diameter 
of  that  semicircle  as  an  axis  has  a  definite  limit  if  the  num- 
ber of  chords  in  the  chain  is  increased  indefinitely. 

194.  Since  we  cannot  measure  the  space  inclosed  by  any 
sphere  in  terms  of  cubic  units,  we  shall  have  to  define  what 
is  meant  by  the  volume  of  a  sphere. 

The  volume  of  a  sphere  is  defined  as  the  limit  of  the 
volumes  of  the  solids  inclosed  by  the  surfaces  generated 
when  a  series  of  chains  of  equal  chords  inscribed  in  a  semi- 
circle revolves  about  the  diameter  of  that  semicircle  as  an 
axis,  as  the  number  of  chords  in  the  chain  is  increased 
indefinitely. 

196.  Therefore,  since  the  volume  inclosed  by  the  surface 
generated  when  a  chain  of  equal  chords  inscribed  in  a 
semicircle  revolves  about  the  diameter  of  that  semicircle 
as  an  axis  is  always  the  product  of  the  area  of  that  surface 
and  one-third  the  radius,  we  will  assume 

THEOREM  130.  The  volume  of  a  sphere  is  the  product 
of  the  area  of  its  surface  and  one-third  its  radius. 

If  r  represents  the  radius  of  the  sphere,  V  its  volume,  and 
A  the  area  of  its  surface,  we  have 

or  V  =  K 


152  SOLID   GEOMETRY 

Ex.  1.  Find   the  volume  of  a  sphere  whose  radius  is  2  in.; 
3  in.;  7  in. 

Ex.  2.  How  many  iron  balls  2  in.  in  diameter  can  be  made 
from  one  15  in.  in  diameter? 

Ex.  3.  Find  the  volume  of  a  spherical  shell 
3/2  in.  thick  if  its  outer  diameter  is  6  in. 

Ex.  4.  Make  a  formula  for  the  volume  of  a 
spherical  shell  if  its  thickness  is  h  and  its  outer 
diameter  d. 

Ex.  5.  A  solid  is  in  the  form  of  a  right  circular 
cone  with  hemisphere  on  its  base  (Fig.  173).     If 
the  diameter  of  the  base  is  6  in.  and  the  slant  height  of  the  cone 
8  in.,  find  the  total  area  and  the  volume. 

VOLUMES    OF   SPHERICAL   SECTORS 
196.  The  solid  generated  by  a  circular  sector  revolving 
about  a  diameter  that  does  not  cross  the  sector  is  called  a 
spherical  sector  (Fig.  174), 

COR.     The  volume  of  a  spherical  sector 
is  the  product  of  the  area  of  the  zone  which 
forms  its  base  and  one-third  of  the  radius    \ 
of  the  sphere. 

The  corollary  may  be  verified  as  follows:          FIG.  174 
Describe  on  the  arc  of  the  circular  sector  any  number 
of  equal  chords  (Fig.  175).    Join  the  points  of 
division   on  the   arc   with  the  center  of  the 
circle,  thus  inscribing  in   the  sector  a  series 
of    congruent     triangles.     Find    the    volume 
generated  when  the  series  of  triangles  revolves 
about  the  diameter,  and  proceed  as  in  the      ^ 
discussion  for  Th.  130. 

If  r  represents  the  radius  of  the  sphere,  A  the  area  of  the 
zone,  and  V  the  volume  of  the  spherical  sector,  we  have 


Since  A  =  2  wrh  ,          V  = 
where  h  is  the  altitude  of  the  zone. 


AREAS  AND   VOLUMES 


153 


VOLUMES   OF   SPHERES  BY   CAVALIERI'S   THEOREM 
197.  THEOREM  131.    Any  section  of  a  sphere  is  con- 
stantly equal  to  that  of  the  solid  between  the  circumscribed 
cylinder  of  revolution  and  a  double   cone   of  revolution 
inscribed  in  the  cylinder. 


CC  D 
YY  X 


BB 


FIG.  176 


Hypothesis:  0  is  any  sphere,  ABCD  the  circumscribed 
cylinder  of  revolution,  and  AOB-COD  a  double  cone  of 
revolution  inscribed  in  the  cylinder.  XY  is  a  plane  cutting 
the  three  solids.  KW,  KZt  and  KY  are  the  radii  of  the 
circles  cut  from  the  cone,  the  sphere,  and  the  cylinder. 

Conclusion:    Area  of  circle  KZ  =  area  of  ring  WY. 

Outline  of  proof: 

Area  of  ring  WY  =  irKY*-  wKW* 


=  ir(KY  -  KO)  (prove  KO  =  KW) 
=  ir(OZ-KO)  (prove  KY  =  OZ) 

_  2 

=  irKZ  =  area  of  circle  KZ. 

THEOREM  130.     The  volume  of  the  sphere  is  tiwr3. 
Suggestion.     It  follows  at  once  from  Cavalieri's  Theorem  that  the 
volume  of  the  sphere  is  the  difference  between  the  volumes  of  the 
circumscribed  cylinder  of  revolution  and  a  double  cone  inscribed  in 
the  cylinder. 

/.  vol.  of  sphere  =7rr2/*-i^7rr2A  =  %7rr2A.     But  h=2r. 

/.   F 
11 


154  SOLID   GEOMETRY 

VOLUMES   OF   SPHERICAL   WEDGES  AND    SPHERICAL 
PYRAMIDS 

198.  A  solid  bounded  by  a  lune  and  the  plane  of  its 
sides  is  called  a  spherical  wedge. 

A  solid 'bounded  by  a  spherical  polygon  and  the  planes 
of  its  sides  is  called  a  spherical  pyramid. 

The  volume  of  a  spherical  wedge  or  of  a  spherical  pyra- 
mid may  be  obtained  from  the  following  assumption: 

As.  31.  The  ratio  of  the  volume  of  a  spherical  wedge 
or  of  a  spherical  pyramid  to  the  volume  of  the  sphere  equals 
the  ratio  of  the  area  of  the  base  of  the  wedge  or  of  the 
pyramid  to  the  area  of  the  sphere. 

If  A  is  the  area  of  the  base  of  the  wedge  or  of  the  pyramid 
and  V  its  volume, 


Since  the  area  of  a  lune  is  ^ 


COR.  I.  The  volume  of  the  spherical  wedge  is  j£=  Trr3. 

2Tv 

Since  the  area  of  a  spherical  triangle  is  T^nr2, 

loU 

COR.  II.  The  volume  of  the  spherical  pyramid  is  =^r  Trr3. 


Ex.  1.  Find  the  volume  of  a  spherical  wedge  cut  from  a  sphere 
whose  radius  is  6  in.  if  the  angle  of  the  lune  is  72°. 

Ex.  2.  Find  the  volume  of  a  spherical  pyramid  whose  base  is 
a  spherical  triangle  with  angles  156°,  94°,  and  128°,  cut  from  a 
sphere  8  in.  in  diameter. 

A  spherical  cone  is  something  like  a  spherical  pyramid. 
Its  base  is  a  zone  of  one  base. 

Ex.  3.  Make  a  formula  for  the  volume  of  a  spherical  cone. 
As.  31  holds. 


AREAS  AND   VOLUMES 
VOLUMES   OF  SPHERICAL   SEGMENTS 


155 


199.  A  portion  of  a  sphere  included  between  two  parallel 
planes  is  a  spherical  segment. 

THEOREM  132.  If  a  and  b  represent  the  radii  of  the 
bases  of  a  spherical  segment,  h  its  altitude,  r  the  radius  of 
the  sphere,  and  V  the  volume  of  the  segment,  then 


FIG.  177 
Analysis: 

I.  Find   the    volume    of    the    frustum    generated    by 

CAHBD  and  add  the  volume  generated  by  AKBH. 
II.  To  find  the  volume  generated  by  AKBH,  subtract 
the  volume  generated  by  &ABO  from  the  volume 
generated  by  the  circular  sector  AKBO. 

Outline  of  proof: 
I.  Vol.  of  sector  =  %7rr2/t. 

II.  Vol.  of  AABO  =  YzOH  •  area  AB. 

But  area  AB  =  2?r  •  OH  •  h, 


III. 


vol.  of 
vol.  of 


=  2/3Trh(r*-OH*) 


y6irhAB2. 


From 

/.vol.  of 
From  &ABE,  AB2  =  h*+(a-b)\ 

.'.  vol.  AKBH  =  M7rh(hz+a?+bz-2ab). 


IV. 
V. 


Vol.  of 
/.  vol.  of  sph.  seg. 


156  SOLID   GEOMETRY 

SUMMARY.  AND   SUPPLEMENTARY  EXERCISES 
FORMULAE   OBTAINED 

200.  NOTE.  In  the  formulae  below  L  =  lateral  area,  A  =  total 
area,  V  =  volume,  b  —  base,  r=  radius,  h  =  height,  /=  slant  height, 
p  =  perimeter,  e  =  edge  of  prisms,  elements  of  cylinders,  spherical 
excess  of  spherical  triangles,  a  =  angle. 

A.  PRISMS: 

L  (any  prism)  =  p  of  rt.  sec.  •  e. 
L  (rt.  prisms)  =  p  of  base  •  e. 
V  (any  prism)  =?bh. 

B.  PYRAMIDS: 

L  (regular)  =Y^(p  of  base)  •  /. 
V  (any  pyramid) 

C.  PRISMATOIDS  : 


D.  FRUSTUMS  OF  PYRAMIDS: 

L  (regular)  =K/  (sum  of  p  of  bases). 


E.  CYLINDERS: 

L  (circular)  =p  of  rt.  sec.  •  e 
L  (rt.  circular)  =2irrh  . 
A  (rt.  circular)  =2  irr(r-\-h}. 
V  (circular  base)  =  7rr2h. 

F.  CONES: 

L  (rt.  circular)  =  irrl. 

A  (rt.  circular)  =  irr(r+l). 

V  (circular  base) 

G.  FRUSTUMS  OF  CONES: 

L  (rt.  circular  )  =  7r/ 
V  (circular  base)  = 

H.  Area  generated  by  a  segment  revolving  about  an  axis 
=  projection  of  segment  on  axis  times  2ir  perpen- 
dicular from  mid-point  of  segment  to  axis. 


AREAS  AND  VOLUMES  157 

I.  Volume  of  solid  formed  by  revolving  a  triangle  about  an 
axis  =  Yz  area  generated  by  revolving  one  side  about 
a  fixed  vertex  times  the  corresponding  altitude. 

J.  SPHERES: 

A  of  sphere  =  47^. 
A  of  zone  =  2?rrk 


A  of  lune  =  2a  sph.  deg.  =  -^-  . 

A  of  sph.  triangle  =  e  sph.  deg.  =7^77. 

loU 

V  of  sphere  =  H  irr3. 
V  of  sph.  sector  =  H  irr2h. 
V  of  sph.  wedge  =  ^TTT  TIT*. 
V  of  sph.  pyramid  =  TTTT  Trr3. 
V  of  sph.  segment  =  %wh  (a2  +62)+^7r/i3  when  a 
and  6  are  the  radii  of  the  bases  of  the  segment. 

NUMERICAL   PROBLEMS 

201.  1.  Find  the  number  of  cubic  yards  of  concrete  required 
for  the  foundation  walls  of  a  house  24  ft.  X  20  ft.  The  walls  are 
to  be  10  in.  thick  and  8  ft.  high.  No  deductions  for  windows. 

2.  Two  congruent  regular  square  pyramids  stand  on  opposite 
sides  of  the  same  base.     The  distance  between  their  vertices  is 
16  cm.    The  diagonal  of  the  com- 

mon  base   is   16  cm.     Find  the 
total  area  and  the  volume. 

3.  Find  the    volume    of    the 
wedge  shown  in  Fig.   178.     The 

base  A  BCD  is  a  rectangle  10  in.  FlG   178 

X18    in.      XEF   and    YGH    are 

right  sections  and  are  isosceles  triangles.     Use  the  data  given. 
XZ  is  _L  base. 

4.  Solve  the  previous  exercise  by  the  prismatoid  formula. 

5.  Find  the  formula  for  the  volume  of  a  wedge  if  the  dimensions 
of  the  base  are  a  and  b,  the  altitude  is  h,  and  the  edge  XY  (Fig.  178) 
is  c. 


158 


SOLID   GEOMETRY 


6.  Find  the  volume  of  a  regular  octahedron  if  one  edge  is  6  in.; 
if  one  edge  is  E. 

7.  Find  the  volume  of  a  regular  tetrahedron  if  one  edge  is  6  in. ; 
if  one  edge  is  E. 

8.  A  berry  box  is  in  the  form  of  a  frustum  of  a  regular  square 
pryamid  5  in.  square  at  the  top  and  4^  in.  square  at  the  bottom. 
What  should  be  the  depth  of  the  box  if  it  holds  a  quart?    Use 
1  dry  qt.  =  67.2  cu.  in. 

9.  A  pail  is  in  the  form  of  a  frustum  of  a  right  circular  cone 
the  radii  of  whose  bases  are  12  in.  and  10  in.  respectively.    Find 
the  depth  of  the  pail  if  it  holds  2^  gal.     1  gal  =  231  cu.  in. 

10.  Fig.   179   shows    a 
cylindrical  tank  partly  rilled 
with  water.    The  tank  is 
6  ft.  long  and  4  ft.  in  diam- 
eter.   If  the  greatest  depth 
of  the  water  is  3  ft.,  find  the 
number  of  gallons  of  water  - 
in  the  tank. 


FIG.  179 


NOTE.    Problems  like  Ex.  10  cannot  be  solved  without  trigonometry 
except  in  special  cases. 

11.  Find  the  total  area  and  the  volume  of  the 
solid  formed   by   revolving  the  square  A  BCD 
(Fig.   180)  about  line  /,  if  /  is  JL  diagonal  AC 
at  C.    Use  data  given. 

12.  Find  the  total  area  and  the  volume  of  the 
solid  formed  by  revolving  the  equilateral  triangle 
ABC  (Fig.  181)  about  line  /,  if  /  is  JL  the  altitude 
DB  at  point  B.    Use  data  given. 

13.  How  many  shot  ¥%  in.  in  diameter  can 
be  made  from  a  cylindrical  bar  10  in.  long  and 
^8  in.  in  diameter? 

14.  Find  the  weight  of  a  spherical  shell  of  iron 
if  its  outside  diameter  is  6  in.  and  it  is  %  of  an 
inch  thick.     Use  1  cu.  in.  of  iron  weighs  0 . 28  Ib. 

15.  Given  a 'cube  whose  side  is  2  in.    Find  the  area  and  the 
volume  of  (1)  the  inscribed  sphere;  (2)  the  circumscribed  sphere. 


FIG.  181 


AREAS  AND  VOLUMES  159 

16.  Find  the  radius  of  a  sphere  that  is  equal  in  volume  to  two 
spheres  whose  radii  are  (1)  2  in.  and  5  in.  respectively;  (2)  .Rand 
r  respectively. 

17.  Find  the  area  of  a  trirectangular  spherical  triangle  on  a 
sphere  whose  radius  is  18  in. 

18.  If  the  earth  is  a  perfect  sphere,  prove  that  one-half  of  its 
surface  lies  within  30°  of  the  equator. 

19.  What  is  the  area  in  spherical  degrees  of  a  spherical  triangle 
if  the  sides  of  its  polar  are  72°,  96°,  and  40°? 

20.  Find  each  angle  of  an  equilateral  spherical  triangle  if  its 
area  is  one-third  the  area  of  the  sphere. 

21.  Find  the  angle  of  a  lune  if  its  area  is  two-ninths  the  area  of 
the  sphere. 

22.  Find  the  angle  of  a  lune  that  is  equivalent  to  a  spherical 
triangle  each  angle  of  which  is  84°. 

23.  The  area  of  a  lune  is  154  sq.  in.     Find  its  angle  if  the 
radius  of  the  sphere  is  9  in. 

24.  The  area  of  a  lune  whose  angle  is  48°  is  54  sq.  in.    Find  the 
radius  and  the  area  of  the  sphere. 

25.  How  many  degrees  are  there  in  the  angle  of  a  lune  if  its 
area  is  equal  to  Y$  the  area  of  a  great  circle  of  the  sphere. 

26.  The  polar  distance  of  a  small  circle 
on  a  sphere  is  60°.    The  radius  of  the  sphere 
is  15  in.Find  the  radius  of  the  circle,  the  dis- 
tance of  the  plane  of  the  circle  from  the  center 
of  the  sphere,  and  the  area  of  the  zone  of  one 
base  cut  off  by  the  circle. 

27.  The  circumference  of  a  great  circle  of  a  sphere 
is  8ft.  Find  the  radius  of  the  sphere;  also  its  area  and 
volume. 

28.  Given  a  sphere  whose  center  is  0,  and  A  a  point 
of  light  outside  of  sphere  0.    Find  the  area  illuminated 
by  A.    Let  the  radius  of  the  sphere  be  10  ft.,  and 
OA  be  15  ft.  (Fig.  182). 

29.  Fig.  183  shows  a  post  which  is  capped  by  a 
portion  of  a  sphere.     The  height  of  the  post  without      FIG.  183 
the  sphere  is  3  ft.    Its  diameter  is  15  in.    The  radius  of  the  sphere 
is  15  in.     Find  the  total  volume. 


160  SOLID   GEOMETRY 

30.  The  water  tank  shown  in  Fig.  184  consists  of  a  cylinder 
with   a  hemisphere  below  and  a  cone  above.     The  diameter  is 
20  ft.  and  the  height  of  the  cylinder  is  32  ft.     The  conical 

roof  is  10  ft.  high.  Find  the  capacity  of  the  tank  in 
gallons.  231  cu.  in.  =  l  gal.  How  much  paint  will  be 
required  to  paint  the  outside  of  the  tank,  allowing  one 
gallon  of  paint  for  500  sq.  ft.  of  surface? 

31.  A  hemispherical  cap  of  aluminum  is  3  in.  in  diam- 
eter.    Find  the  diameter  of  the  blank  from  which  it  is    N — / 

FIG.  184 
pressed. 

NOTE.  The  blank  is  a  circular  piece  of  metal  cut  from  a  flat  sheet. 
This  is  pressed  into  the  desired  form  by  means  of  a  die.  Assume  that 
the  area  of  the  blank  is  equal  to  the  area  of  the  finished  article. 

32.  Find  the  diameter  of  the  blank  if  the  cap  in  the  preceding 
exercise  has  a  flat  ring  ^2  in.  around  it. 

33.  A  spherical  shell  of  iron  whose  outer  diameter  is  1  ft.  is 
filled  with  lead.     Find  the  thickness  of  the  iron  if  the  filled  shell 
weighs  twice  the  unfilled  shell.     A  cubic  inch  of  iron  weighs  4.2  oz. ; 
a  cubic  inch  of  lead  weighs  6.6  oz. 

34.  What  is  the  ratio  of  the  surface  of  a  sphere  to  the  entire 
surface  of  its  hemisphere? 

35.  A  plane  is  drawn  tangent  to  the  inner  of  two  concentric 
spheres  8.  in.  and  12  in.  in  diameter.     Find  the  circumference  of 
the  circle  cut  out  on  the  outer  sphere. 

36.  What  is  the  altitude  of  a  zone  of  a  sphere  which  equals  a 
trirect angular  triangle  in  area? 

37.  The  diameter  of  a  right  circular  cylinder  equals  its  height. 
Find  its  dimensions  if  its  capacity  is  one  gallon.     231  cu.  in.  =  1  gal. 

38.  The  area  of  a  sphere  is  equal  to  the  lateral  area  of  the 
circumscribed  cylinder  of  revolution. 

39.  With  the  compasses  open  4  in.  a  circle  is  drawn  on  a  12-in. 
globe.     Find  the  circumference  of  the  circle,  and.  the  area  of  the 
zone  of  one  base  cut  off. 

40.  A  cylindrical  bore  is  made  through  a  sphere.     If  the  radius 
of  the  sphere  is  6  in.,  and  the  diameter  of  the  bore  6  in.,  find  the 
entire  area  of  the  part  removed. 


AREAS  AND   VOLUMES 


161 


EXERCISES   INVOLVING   PROOFS 

202.  1.  The  volume  of  a  triangular  prism  is  the  product  of 
one  lateral  face  and  one-half  the  distance  of  that  face  from  the 
opposite  edge. 

2.  The  volume  of  a  regular  prism  is  the  product  of  the  lateral 
area  and  one-half  the  altitude  of  the  base. 

3.  Any   plane    passed   through 
the  center  of  a  parallelepiped  di- 
vides it  into  two  equivalent  parts. 

4.  A  truncated  triangular  prism 
is  equivalent    to    three    pyramids 
whose  common  base  is  the  lower 
base  of  the  prism  and  whose  ver- 
tices are  the  vertices  of  the  upper 

base  (Fig.  185).  FlG  185  . 

.  1  mily  sis: 

I.  ABC-DEF  =  E-ABC+E-A  CD+E-CDF. 
II.  E-ABC  is  one  of  the  required  pyramids. 

III.  Prove  E-A  CD  =  D-ABCby  proving  both  equivalent  to  B-A  CD. 

IV.  Prove  E-CDF=  F-ABCby  proving  F-ABC  =  B-A  CF  =  B-CFD 

=  E-CFD. 

5.  The  volume  of  a  truncated  right  triangular  prism  is  the 
product,  of  the  area  of  its  base  and  one-third  the  sum  of  the  lateral 
edges.     (See  Ex.  4.) 

6.  The  volume  of  any  truncated  triangular  prism  is  the  prod- 
uct of  the  area  of  a  right  section  and  one-third  the  sum  of  its 
lateral  edges. 

Suggestion.     The  right  section  divides  it  into  two 
truncated  right  prisms. 

7.  The  area  of  a  zone  of  one  base  is  equal  to  the 
area  of  a  circle  whose  radius  is  the  generating  arc. 

Suggestion.     Since  the  area  of  the  zone  generated 
by  AB  (Fig.  186)  is  2irrBE,  and  the  area  of  the  circle 


_ 

whose  radius  is  AB  is  wAB 
BCBE- 


_ 

prove  that  2irrBE  =  irAB  . 


prove 


CHAPTER  V 
SIMILARITY  AND  SYMMETRY 

SIMILARITY 
TESTS   FOR   SIMILAR   SOLIDS 

203.  Two  polyhedrons  are  said  to  be  similar  if  they  have 
the  same  number  of  faces  respectively  similar,  and  similarly 
placed,  and  their  corresponding  polyhedral  angles  equal. 

For  convenience,  corresponding  parts  will  be  lettered 
alike;   for  example,  AB  and  A 'B'  are  corresponding  edges. 

204.  THEOREM  133.    Two   tetrahedrons   are    similar   if 
three  faces  meeting  at  a  common  vertex  in  one  are  respec- 
tively similar  to  three  faces  meeting  at  a  common  vertex 
of  the  other,  and  arranged  in  the  same  order. 


FIG.  187 

Hypothesis:     T  and  T'  are  two  tetrahedrons  with  ADB  ~ 
A'D'B',  BDC**B'D'Cf,  and  ADC**A'D'C'. 
Conclusion:     T  co  T'. 

I.  To  prove  T^T',  prove  AABC«>  &A'B'C',  and 
the  trihedral  angles  at  A,  B,  C,  and  D  equal 
respectively  to  the  trihedral  angles  at  Af,  B',  C', 
and  D'. 

162 


SIMILARITY  AND   SYMMETRY  163 

Analysis: 

II.  To  prove  the  trihedral  angle  at  A  equal  to  the  tri- 
hedral angle  at  A',  prove  the  face  angles  of  one 
equal  to  the  face  angles  of  the  other,  and  arranged 
in  the  same  order. 

III.   .'.   prove  AABC^AA'B'C. 

AB  _  EC      CA 
eA7B)~BfCf"CrAf' 

206.  If  the  segments  that  join  the  vertices  of  a  given 
polyhedron  with  a  given  point  are  divided  in  the  same 
ratio  from  the  given  point,  and  if  the  points  of  division  are 
joined  in  the  same  order  as  the  vertices  of  the  given  poly- 
hedron, the  polyhedron  so  formed  and  the  given  polyhedron 
are  radially  placed.  The  given  point  may  be  called  the 
radial  center. 


FIG.  188 

The  radial  center  of  the  two  radially  placed  polyhedrons 
may  be  within  or  without  the  polyhedrons.  If  it  is  without 
them,  the  two  polyhedrons  may  be  on  the  same  or  on 
opposite  sides  of  the  center.  Let  the  pupil  draw  figures 
for  each  case. 

206.  THEOREM  134.  Two  dihedral  angles  are  equal  if 
the  faces  of  one  are  parallel  to  the  faces  of  the  other  and 
extend  in  the  same  direction  from  their  edges. 

For  proof  see  suggestion  to  Th.  24,  p.  23. 


164 


SOLID   GEOMETRY 


207.  THEOREM  135.  If  two  polyhedrons  are  radially 
placed,  and  are  on  the  same  side  of  the  radial  center,  the 
polyhedrons  are  similar. 

K  G 


FIG.  189 

Hypothesis:     P   and   Pf   are   two   polyhedrons   radially 
placed  on  the  same. side  of  the  radical  center  0. 
Conclusion:     P<»P'. 
Analysis: 
I.  To  prove  P^P',  prove 

(1)  The  faces  of  P  similar  to  the  corresponding 

faces  of  P'. 

(2)  The  polyhedral  angles  of  P  equal  to  the  cor- 

responding polyhedral  angles  of  P' . 

(3)  The  parts  are  similarly  placed. 

II.  To  prove  face  ABFE  ~  face  A'B'F'E',  prove   .    .    .    . 

III.  To  prove  the  polyhedral  angle  at  B  equal  to  the 

polyhedral  angle  at  Bf,  prove  that  the  face  angles 
and  the  dihedral  angles  of  one  are  equal  respec- 
tively to  the  corresponding  parts  of  the  other. 

IV.  Verify  that  the  parts  are  arranged  in  the  same  order. 

When  two  polyhedrons  are  radially  placed,  the  radial 
center  is  called  the  center  of  similitude. 

The  ratio  of  the  distances  of  corresponding  vertices  from 
the  center  of  similitude  is  called  the  ratio  of  similitude. 


SIMILARITY  AND   SYMMETRY 


165 


208.  THEOREM    136.    If   two   polyhedrons   are   radially 
placed,  with  the  center  of  similitude  within  the  polyhedrons, 
the  polyhedrons  are  similar. 

209.  THEOREM    137.    If   two   polyhedrons   are  radially 
placed,  and  are  on  opposite  sides  of  the  radial  center,  the 
faces  of  one  are  similar  to  the  corresponding  faces  of  the 
other,  the  polyhedral  angles  of  one  are  equal  to  the  corre- 
sponding polyhedral  angles  of  the  other,  but  the  parts  are 
arranged  in  the  reverse  order. 

210.  Two  polyhedrons  in  the  position  called  for  in  Th. 
135  are  said  to  be  in  the  direct  radial  position  and  are  said 
to  be  directly  similar.     If,  however,  the  two  polyhedrons 
are  in  the  position  called  for  in  Th.  137,  they  are  said  to  be 
in  the  inverse  radial  position  and  are  said  to  be  inversely 
similar. 


FIG.  190 

Although  the  proofs  are  too  difficult  for  this  book,  the 
statements  above  apply  to  all  kinds  of  solids.  These 
definitions  may  be  used  as  fundamental  definitions  of 
similar  solids  (Figs.  190  and  191). 


FIG.  191 


166 


SOLID  GEOMETRY 


PROPERTIES   OF   SIMILAR   SOLIDS 

211.  As.  32.  Two  solids  similar  to  a  third  are  similar 
to  each  other. 

THEOREM  138.  Two  similar  polyhedrons  may  be  placed 
in  the  direct  radial  position. 

p 

G 


FIG.  192 

Hypothesis:    P  and  Q  are  two  similar  polyhedrons. 
Conclusion:    P  and  Q  can  be  placed  in  the  direct  radial 
position. 

Analysis  and  construction: 
I.  It  is  necessary  to  prove  Q  congruent  to  a  polyhedron 

that  is  in  the  direct  radial  position  with  P. 
II.   /.   draw  rays  to  any  point  0,  and  construct  Q'  in 

OA       AB 
direct  radial  position  with  P  so  that  -~-r>  =  ./p/ . 

C//I          './i   JD 

III.  To  prove  Q^Q',  prove  the  polyhedral  angles  of  Q 

equal  to  the  corresponding  polyhedral  angles  of 
Q',  and  the  faces  of  Q  congruent  to  the  corre- 
sponding faces  of  Q' . 

IV.  To  prove  the  corresponding  polyhedral  angles  equal, 

prove  Q^Q'. 

V.  To  prove   A^F^  ^A'B'F'E',  prove  Ai5iFi£i~ 
A'B'F'E'  and  A^^A'B'. 

AB      OA      AB 
VI.   ••• 


SIMILARITY  AND  SYMMETRY 


167 


212.  THEOREM   139.     Two  similar  polyhedrons  can  be 
divided  into  the  same  number  of  tetrahedrons  similar  each 
to  each  and  similarly  placed. 

Suggestion.  Pass  a  plane  through  ACF  and  AiC\Fi  in  P  and  Q 
(Fig.  192)  and  show  that  the  tetrahedrons  cut  from  P  and  Q  are 
similar.  Continue  in  this  way  until  P  and  Q  are  completely  divided 
into  tetrahedrons. 

213.  THEOREM    140.     If   two   polyhedrons   are    similar, 
the  ratio  of  any  two  corresponding  edges  equals  the  ratio 
of  similitude. 

Suggestion.  Place  the  two  similar  polyhedrons  in  the  direct  radial 
position. 

214.  NOTE.     It  is  also  true  that  if  two  polyhedrons  are  similar 
the  ratio  of  any  two  corresponding  segments  equals  the  ratio  of  simili- 
tude.    We  will  prove  here  only  one  special  case  under  this  theorem. 
This  special  case  is  Th.  141  and  is  needed  in  the  proof  to  Th.  143.    Let 
the  pupil  name  other  special  cases. 

THEOREM  141.  If  two  tetrahedrons  are  similar,  the  ratio 
of  corresponding  altitudes  equals  the  ratio  of  similitude. 


D' 


FIG.  193 

Analysis  and  construction: 

T    _  DX      AB  DX      AD      AB 

I.  To  prove £7^=^,,  prove  g^ —--j^,. 

II.  To  prove  7^77  = -p^-p, ,  place  T'  on  T  so  that  the  tri- 

U  A       LJ  A 

hedral  angle  D'  fits  upon  the  trihedral  angle  D  and 
prove  (1)  A'B'C'  parallel  to  ABC;  (2)  D'X'  falls  on 
DX. 


168 


SOLID   GEOMETRY 


215.  THEOREM  142.    If  two  polyhedrons  are  similar,  the 
ratio  of  the  areas  of  corresponding  faces  equals  the  square 
of  the  ratio  of  similitude. 

Show^that  this  reduces  to  a  plane  geometry  theorem. 

216.  THEOREM    143.     If  two   tetrahedrons   are   similar, 
the  ratio  of  the  volumes  equals  the  cube  of  the  ratio  of 
similitude. 

D 


D' 


B' 


Hypothesis:    D-ABC ™ D-A'B'C',  b   and   b'  represent 
the  bases,  a  and  a'  the  altitudes,  and  V  and  V  the  volumes. 


Conclusion:     —  = 


V    A'B' 

•    Analysis:     To  prove  —  ==    ~,  prove  (1)  TT/  =  -777  J  (2) 

V    A'B' 

a_AB  b.^AB* 

a'  'A'B-     )  b,    A&- 

Outline  of  proof: 


V  =y3ab    _>ab 
'  V        a'b'     a'b1 


b      AB' 


*  a'    A'B 


A'B' 


ab  _  AB 
a'b' 


A'B 


73 


V    A'B' 


SIMILARITY  AND   SYMMETRY 


169 


THEOREM  144.    If  two  polyhedrons  are  similar,  the  ratio 
of  the  volumes  is  equal  to  the  cube  of  the  ratio  of  similitude. 


FIG.  195 


Hypothesis:    P^P'.     V  and  V  represent  the  volumes. 

4 

y 

Conclusion:      —  =  

V    A'B'* 

Analysis  and  construction: 


I.  To  prove  


V     AB 


,  divide  P  and  P'  into  similar 


V    A'B'S 
tetrahedrons  T\  and  7Y,  T2  and  7Y,  T3  and  TV, 

ri+r2+r3,  etc.     AB* 


etc.,  and  prove 


Ti     AB     72 
II.    /.  prove  —  = r.  —  = 


TY+TY+Ty,  etc.     ^7^73* 
,  etc. 


Suggestion.     See  Th.  124,  P/ane  Geometry. 

Ex.  1.  Are  regular  polyhedrons  of  the  same  number  of  faces 
similar?  Why? 

Ex.  2.  The  bases  of  two  similar  pyramids  are  in  the  ratio  of 
4 : 9.  What  is  the  ratio  of  their  volumes?  Of  their  altitudes? 

Ex.  3.  The  dimensions  of  a  box  are  6,  8,  and  12.  Find  the 
dimensions  of  a  similar  box  that  will  hold  twice  as  much. 

217.  Two  cylinders  or  two  cones  of  revolution  are  said 
to  be  similar  if  they  are  generated  by  the  revolution  of 
similar  figures  revolving  about  corresponding  sides. 
12 


170 


SOLID   GEOMETRY 


THEOREM  145.  The  lateral  areas  or  the  total  areas  of 
two  similar  cylinders  of  revolution  have  the  same  ratio  as 
the  squares  of  their  radii  or  the  squares  of  their  altitudes. 


h' 


C' 


FIG.  196 

Hypothesis:  C  and  C'  are  two  similar  cylinders  of  revo- 
lution; r  and  r'  represent  the  radii,  h  and  h'  the  altitudes, 
L  and  L'  the  lateral  areas,  and  A  and  A'  the  total  areas. 

Conclusion: 

T    ~L——  =  —• 

T'      v't       h'% 
J-/          /  i  If 

Outline  of  proof: 
T         L  _  2irrh  _  rh 
"  V~2^7kf~7h'' 

r      h  rh      r2      h2 

2.  Since  p-p,  /.  -^,=-5=— 

L_r2_^2 
^   ••  I/-^-^2- 

A       27rr(^+r)        r(fe+r) 
1X-  *•  T>  =  ; 


A'    27rr/(/t/+r/)     r/(/t'+? 
r      /z.          r      h+r 


~h"    '  r'-h'+r' 
r(h+r)       r2     h2 . 


rf(hf+rf)  .r'2    h'2 

4'    '*•  Z>  =  r75==Sr2' 
NOTE  for  II.  2.     See  P/awe  Geometry,  Th.  124. 


SIMILARITY  AND   SYMMETRY 


171 


THEOREM  146.  The  lateral  areas  or  the  total  areas  of 
two  similar  cones  of  revolution  have  the  same  ratio  as  the 
squares  of  the  radii  or  the  squares  of  the  altitudes. 


FIG.  19 


THEOREM  147.  The  volumes  of  two  similar  cylinders 
of  revolution  have  the  same  ratio  as  the  cubes  of  the  radii 
or  the  cubes  of  the  altitudes  (Fig.  196). 

Outline  of  proof: 


'  V     irrW-     r'2h'2 
II.  Sincep-p'  /.  pj-j^andpi-p- 

III.  .*.    —  =  —  =  —  . 
r^2h^     h^     T^ 

iv   •  ?-,*-». 

"  y    r'3    7/3 

THEOREM  148.  The  volumes  of  two  similar  cones  of 
revolution  have  the  same  ratio  as  the  cubes  of  the  radii  or 
the  cubes  of  the  altitudes  (Fig.  197). 

218.  Any  two  spheres  are  similar. 

THEOREM  149.  The  areas  of  two  spheres  have  the 
same  ratio  as  the  squares  of  their  radii. 

THEOREM  150.  The  volumes  of  two  spheres  have  the 
same  ratio  as  the  cubes  of  their  radii. 

Exercise.  The  radii  of  two  spheres  are  in  the  ratio  of  2:5. 
The  sum  of  their  volumes  is  3994  cu.  in.  Find  the  volumes. 


172  SOLID   GEOMETRY 

219.  The   fundamental    definition   for   similar   solids   is 
given  in  §210.     The  following  facts  are  true  for  all  similar 
solids  regardless  of  shape: 

I.  The  ratio  of  corresponding  segments  is  equal  to  the 
ratio  of  similitude. 

II.  The   ratio   of   the   areas   of   corresponding   surfaces 
equals  the  square  of  the  ratio  of  similitude. 

III.  The  ratio  of  the  volumes  is  equal  to  the  cube  of  the 
ratio  of  similitude. 

Only  special  cases  have  been  proved  in  this  book. 

SYMMETRY 

220.  We  have  defined  two  polyhedrons  as  symmetric  if 
the  parts  of  one  are  equal  respectively  to  the  parts  of  the 
other,  but  arranged  in  the  reverse  order. 

THEOREM  151.  If  two  polyhedrons  are  in  the  inverse 
radial  position,  and  the  ratio  of  similitude  is  one,  the  poly- 
hedrons are  symmetric  with  respect  to  the  center  of  simili- 
tude. 


FIG.  198 

Analysis:  To  prove  that  P  and  P'  are  symmetric,  prove 
that  the  parts  of  one  are  equal  respectively  to  the  parts  of 
the  other,  but  arranged  in  the  reverse  order. 

By  a  method  similar  to  that  used  in  proving  Th.  138,  we 
may  prove  that  if  two  polyhedrons  are  symmetric  they  may 
be  placed  in  the  inverse  radial  position. 


SIMILARITY  AND   SYMMETRY 


173 


FIG.  199 


221.  Two  solids  may  be  so  situated  on  opposite  sides  of 
a  plane  that  the  plane  bisects  at  right  angles  all  segments 
joining  corresponding  points  of  the 

two  solids.  We  can  prove  that 
in  this  case  the  two  solids  arc 
symmetric.  They  are  said  to  be 
symmetric  with  respect  to  the 
plane.  We  can  prove  also  that  if 
two  solids  are  symmetric  they  can 
be  placed  on  opposite  sides  of  a 
plane  as  indicated  above. 

It  is  to  be  noted  that  two  figures  in  plane  geometry  that 
are  symmetric  with  respect  to  a  point  or  to  a  line  can  be  made 
to  coincide,  but  that  two  solids  that  are  symmetric  with 
respect  to  a  point  or  to  a  plane  cannot  be  made  to  coincide. 

Two  solids  may  be  so  situated  in  regard  to  a  line  that  the 
line  bisects  at  right  angles  all  segments  joining  corresponding 
points  of  the  two  solids.  In  this  case  the  solids  are  said  to 
be  symmetrically  situated  with  regard  to  the  line.  The  two 
solids  are,  however,  congruent,  and  can  be  made  to  coincide. 

MISCELLANEOUS   EXERCISES 

222.  1.  Name  as  many  solids  as  you  can  that  have  a  center  of 
symmetry.     Tell  what  is  the  center  of  symmetry  in  each  case. 

2.  Name  as  many  solids  as  you  can  that  have  a  plane  of  sym- 
metry.    Tell  what  is  the  plane  of  symmetry  in  each  case. 

3.  Give  an  everyday  illustration  of  figures  that  are  symmetric 
with  respect  to  a  plane. 

4.  In  shipping  goods,  which  would  be  more  economical  of  the 
material  of  which  the  packing  boxes  are  made:  .(1)  to  use  two 
boxes  of  the  same  dimensions  or  (2)  to  use  one  box  similar  to  the 
first  that  will  hold  twice  as  much?     Why? 


NOTES  ON  ARITHMETIC  AND   ALGEBRA 

FRACTIONS 

223.  The  following  fundamental  law  of  fractions  under- 
lies all  operations  that  involve  fractions ! 

Multiplying  or  dividing  numerator  and  denominator  of 
a  fraction  by  the  same  number  does  not  alter  the  value  of 
the  fraction. 

A.  The  sum  or  difference  of  two  or  more  fractions  that 
have  a  common  denominator  is  the  sum  or  difference  of  the 
numerators  divided  by  the  common  denominator. 

Two  or  more  fractions  that  have  not  a  common  denomi- 
nator must  be  reduced  to  a  common  denominator  before 
adding  or  subtracting.  To  reduce  fractions  to  a  common 
denominator,  apply  the  fundamental  law  given  above. 

Add  and  subtract  the  following: 

.5,7  a      b  x      to* 

l'  12+F8  3"  b+~c  5'  4~    +4 


"  24  '  36    6  ~  be  '  ac    ab 

B.  The  product  of  two  fractions  is  the  product  of  the 
numerators  divided  by  the  product  of  the  denominators. 
Where  possible,  divide  numerator  and  denominator  by  com- 
mon factors. 

Multiply  the  following: 

3      8  3a    b* 

L16Xl5  2'TX7 

174 


NOTES  ON  ARITHMETIC  AND  ALGEBRA        175 

C.  The  quotient  of  one  fraction  divided  by  a  second  is 

the  product  of  the  first  multiplied  by  the  reciprocal  of  the 
second. 

Divide  the  following: 


6«*  3 

*•  '  Da£7  °- 


L'  20  '  25  '    8&3  '  '  16a&»         2a 

SQUARE  ROOTS 
224.  The  rule  for  square  root  is  based  on  the  algebraic 


formula  (a+b)-  =  a2+2ab+h2.    Notice  that  a2+2a&+62  may 
be  written  a2+6(2a+6).     The  method  is  illustrated  below: 


Illustration  1.     Find    V 694. 563 
694.56,3     |26.3 
4_ 

2(20)  =  40    294 

40+6  =  46    276 

2(260)  =  520       1856 

520+3  =  523      1569 


287 

In  the  illustration  above,  notice: 

(1)  The  number  was  divided  into  periods  of  two  figures 
each,  counting  to  the  left  and  to  the  right  from  the  decimal 
point. 

(2)  The  largest  square  under  6  is  4.     The   4   was   sub- 
tracted from  6  and  the  next  period  annexed.     This  gave  a 
remainder  of  294.     The  square  root  of  4,  or  2,  was  written 
as  the  first  figure  in  the  root. 

(3)  A  zero  was  placed  after  the  2,  making  20.     The  20 
was  doubled,  making  40.     The  40  is  used  as  a  trial  divisor 
for  the  remainder  294.     The  next  figure  of  the  root  is  either 
6  or  7.     The  6  is  added  to  the  40,  making  46.     The  46  is 
multiplied  by  6,  giving  276.     The  276  is  subtracted  from 
294,  leaving  18.     The  next  period  is  annexed,  giving  1856. 

(4)  The  process  above  is  repeated  at  each  step  of  the 
work;  thus,  a  zero  is  placed  after  26  and  the  result  doubled, 
giving  the  520.     The  work  is  then  continued  as  above. 


176  SOLID   GEOMETRY 

In  general  we  may  say:  Annex  a  zero  to  the  part  of  the 
root  already  found  and  double  the  result.  To  this  result 
add  the  next  figure  of  the  root.  Multiply  the  result  by  the 
last  figure  of  the  root  found. 

Show  that  this  statement  may  be  regarded  as  a  trans- 
lation of  6(2a+6)  in  the  formula  (a-f-6)2  =  a2-f-&(2a+&). 

Find  the  square  root  of  the  following: 

1.  1369  4.  106276  7.  6 

2.  3744  5.  3  8.  15 

3.  2304  6.  5  9.  7 

Because  of  the  frequent  occurrence  of  the  square  roots 
of  2  and  3  in  geometry  work,  the  application  of  the  following 
law  should  be  noted: 

225.  The  square  root  of  a  product  is  the  product  of  the 
square  roots  of  the  factors. 

Illustration  2.     36  =  4X9     .'.  V36=  V4X  V9 

This  law  is  used  most  conveniently  for  inexact  square 
roots  when  one  factor  is  a  perfect  square. 

Illustration  3,     18  =  9X2     /.  V 18  =  V9  X  V2  =  3  V2 
Notice  that    V2  occurs  when  the  side  of  a  square  and  a 
diagonal  of  the  square  are  used  in  the  same  exercise. 

Illustration  4.     12  =  4X3. '.      Vl2=  V4  X  V3  =  2  V3 
Notice  that  V3  occurs  when  the  side  of  an  equilateral  tri- 
angle and  its  altitude  occur  in  the  same  exercise. 

Illustration  5.     20  =  4 X  5     /..  V20  =  V4  X  V5  =  2  V5 
The  V5  occurs  in  connection  with  the  regular  decagon  and 
pentagon. 

Find  the   value   of   the   following    correct  to  three   decimal 
places.    Apply  the   law  given   above. 

1.  V8          4.    Vl08      7.    Vl28       10.    V150       13.    V54 

2.  Vl8        5.    V32        8.    V?5        11.    Vl25       14.    V45 

3.  V27        6.    V80        9.    V320      12.    V98        15.    Vl80 


NOTES  ON  ARITHMETIC  AND  ALGEBRA  177 

16.  V20       18.    V48       20.    V50      22.    V288       24.  .V243 

17.  V96      19.    V72      21.    V300    23.    VI62      25.    V242 

226.    The  square  root  of  a  fraction. 

A.  If  the  denominator  is  a  perfect  square  :  Find  the 
square  root  of  the  numerator  and  of  the  denominator  sepa- 
rately and  divide  the  first  result  by  the  second. 


Illustration 


ion  7.    J-  =  K  V§ 

B.  If  the  denominator  is  not  a  perfect  square,  two  methods 
are  suggested: 

(1)  The  fraction  may  be  reduced  to  a  decimal  and  the 
square  root  of  the  result  found. 

(2)  Numerator  and  denominator  may  be  multiplied  by 
some  number  that  will  make  the  denominator  a  perfect 
square  and  method  A  above  used. 

Illustration  8.    To  find  Vlg  either 

1.  Find  square  root  of  .333333+;  or 

2.  Write  H  =  %  and  use  H  V3. 

Find  the  value  of  the  following  correct  to  three  decimal  places: 


1.  Jl  5.JL  9.  Jl 

\25  \5  \5 


13. 


^     6-  \/^      10.  JL      14.  JL 

16  \7  \11  \20 


18 

l"  16 

12  3  27 


_ 
\  ^r 


four 


178  SOLID   GEOMETRY 

EQUATIONS 

227.  The  method  of  solving  linear  equations  is  illustrated 
below  : 

Illustration  1.     Solve  for  x:  ^~  --2  =  4- 


Multiply  both  sides  by  the  L.  C.  M.  of  the 

denominators    ........    3(3*+  1)  -24  =  48  -* 

performing  multiplications  .....     9*  +  3  —  24  =  48—  4x—  4 

Combining  terms  ...........  9#  —  21=44—  4* 

Add  +21  and  +4*  to  each  side    .....    13*  =  65 

Divide  both  sides  by  13  ..........     x=5 

Solve  the  .following  equations: 

^H?=izl  3.     4          3 


3          5  *-5 

2.  2*-7     3+2*  _0  4    3-4*    5-2ac_t      * 

~~T~         4  ~8~        3  ~4 

228.  If  an  equation  contains  both  the  first  and  the  second 
powers  of  the  unknown,  two  methods  of  solution  are 
suggested. 

A.  The  equation  may  be  solved  by  factoring. 
Illustration  2.    Solve  f  or  x :  *2  -  *  =  20 

(*-5)  (*+4)=0 


Notice  that  to  solve  an  equation  by  factoring,  one  member 
of  the  equation  must  be  zero. 

B.  The  equation  may  be  solved  by  completing  the  square. 
Illustration  3.     Solve  for  x:  3x*-5x  =  7     .     .     .     (1) 

Divide  both  sides  by  3 *2~~3"=3~    •     •     •     (2) 

Add  the  square  of  (y£  •  %)  to  each 
side *~+<     *  W*+          -     -     (3) 


NOTES  ON  ARITHMETIC  AND  ALGEBRA         179 
Take  the  square  root  of  each  side 


5 
of  the  equation    ....     *  —  «•=  ± — ^-=±MVl09  .     (4) 

5         10.44 
s-g-i-fl-      ....     (5) 

5  .  10.44  5     10.44 

*=6+~6~  *  =  6         6~ 

=  15.44  5.44 

6  '    6 

=  2.57+  =-.90+ 

Notice  in  (3),  (M  •  2s)2  is  added  to  the  left  side  to  make  the 
left  side  a  perfect  square.  It  is  added  to  the  right  side  to 
preserve  the  balance  of  the  equation.  (>2  •  H)2,  or  2^s,  is 
obtained  by  squaring  half  the  coefficient  of  x.  Notice  that 
in  step  (2)  the  equation  is  divided  by  3  to  make  the  first 
term  x2,  which  is  a  perfect  square. 

Solve  the  following  equations: 

2.  2*2-*  =  15  4.  2*2+5*  =17 

229.     To  solve  a  system  of  equations  consisting  of  two 

equations  containing  two  unknowns,  eliminate  one  of  the 

unknowns  and  solve  the  resulting  equation  for  the  other. 

A.  When  both  equations  are  of  the  first  degree,  eliminate 

by  addition  or  subtraction. 

\5x-4y 


Illustration  4.     Solve  for  x  and  y     ,  _    ,  P    _„« 

5*-4y  =  6.5          (1) 

7^+5^  =  38.25 (2) 

35*-28y  =  45.5 (1)  X  7 

35^+25^  =  191.25 (2)  X  5 

-53y=-145.75 Subtract  the  third 

y  =  2.75  equation  from  the  second 

Notice  that  %  may  be  found  by  multiplying  equation  (1) 
by  5  and  equation  (2)  by  4  and  adding  the  results  or  by 
substituting  2.75  for  y  in  either  equation  (1)  or  (2)  and 
solving  the  result  for  x. 


ISO 


SOLID   GEOMETRY 


B.  When  one  of  the  equations  is  of  the  first  degree  and 
one  of  the  second,  solve  the  first-degree  equation  for  one  oi 
the  unknowns  in  terms  of  the  other  unknown  and  substitute 
in  the  other  equation. 

Illustration  5.     Solve  for  x  and  y:  1*2+2^34 

Solve  (1)  for  x,  x  =  8-y    ...     (3) 

Substitute  8  -  y  for  x  in  (2)         (8  - 1 


3,2-83,4-15  =  0 
(y-5)  (y-3)=0 
y  =  5  and  y  =  3 
To  find  x,  substitute  the  values  of  y  in  (3) : 


x=8-y 
=  8-5 
=  3 

The  solutions  are 


=  8-3 
=  5 


Solve  the  following  systems  for  x  and  y : 


3. 


TABLES 
230.     TABLE   OF   SQUARE   ROOTS 


TABLE  OF  SQUARE  ROOTS  OF  NUMBERS  FROM  0  TO  '99 


0 

1 

2 

O 

4 

5 

6 

7 

8 

9 

0.000 

1.000 

1.414 

1.732 

2.000 

2.236 

2.449 

2.646 

2.828 

3.000 

3.162 

3.317 

3.464 

3.606 

3.742 

3.873 

4.000 

4.123 

4.243 

4.359 

4.472 

4.583 

4.690 

4.796 

4.899 

5.000 

5.099 

5.196 

5.292 

5.385 

5.477 

5.568 

5.657 

5.745 

5.831 

5.916 

6.000 

6.083 

6.164 

6.245 

6.325 

6.403 

6.481 

6.557 

6.633 

6.708 

6.782 

6.856 

6.928 

7.000 

7.071 

7.141 

7.211 

7.280 

7.348 

7.416 

7.483 

7.550 

7.616 

7.681 

7.746 

7.810 

7.874 

7.937 

8.000 

8,062 

8.124 

8.185 

8.246 

8.307 

8.367 

8.426 

8.485 

8.544 

8.602 

8.660 

8.718 

8.775 

8.832 

8.888 

8.944 

9.000 

9.055 

9.110 

9.165 

9.220 

9.274 

9.327 

9.  .381 

9.434 

9.487 

9.539 

9.592 

9.644 

9.695 

9.747 

9.798 

9.849 

9.894 

9.950 

NOTES  ON   ARITHMETIC   AND   ALGEBRA        181 


CUBE  ROOT 

231.  The  rule  for  cube  root  is  based  on  the  algebraic 
formula 


Notice  that 


may  be  written 

a3+6(3a2+3a6+6-) 

The  method  is  illustrated  below: 


Illustration:  Find  ^41523.629 


.  41,523.629  {34.6+ 
27 


3  X(30)2  =  3  X900  =  2700       14523 
3X30X4  =  360 

4'  =     16 


3076 


12304 


2219629 


3  X(340)2  =  3X115600  = 
346800 

3X340X6=  6120 

6*  =  36 


352956 


2117736 


101893 
232.     TABLE   OF   CUBE  ROOTS 


0 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

0 

0 

1 

1.260 

1.442 

1.587 

1.710 

1.817 

1.913 

2.000 

2.080 

1 

2.154 

2.224 

2.289 

2.351 

2.410 

2.466 

2.520 

2,571 

2.621 

2.668 

2 

2.714 

2.759 

2.802 

2.844 

2.885 

2.924 

2.963 

3.000 

3.037 

3.072 

3 

3.107 

3.141 

3.175 

3.208 

3.240 

3.271 

3.302 

3.332 

3.362 

3.391 

4 

3.420 

3.448 

3.476 

3.503 

3.530 

3.557 

3.583 

3.609 

3.634 

3.659 

5 

3.684 

3.708 

3.733 

3.756 

3.780 

3.803 

3.826 

3.849 

3.871 

3.893 

6 

3.915 

3.937 

3.958 

3.979 

4.000 

4.021 

4.041 

4.062 

4.082 

4.102 

7 

4.121 

4.141 

4.160 

4.179 

4.198 

4.217 

4.236 

4.254 

4.273 

4.291 

8 

4.309 

4.327 

4.345 

4.362 

4.380 

4.397 

4.414 

4.431 

4.448 

4.465 

9 

4.481 

4.498 

4.514 

4.531 

4.547 

4.563 

4.579 

4.595 

4.610 

4.626 

182  SOLID   GEOMETRY 

233.     TABLE  OF  SINES,  COSINES,  AND  TANGENTS 


Deg. 

Sine 

Cosine 

Tangent 

Deg. 

Sine 

Cosine 

Tangent 

1 

.017 

.999 

.017 

46 

.719 

.695 

1.036 

2 

.035 

.999 

.035 

47 

.731 

.682 

1.072 

3 

.052 

.999 

.052 

48 

.743 

.669 

1.111 

4 

.070 

.998 

.070 

49 

.755 

.656 

1.150 

5 

.087 

.996 

.087 

50 

.766 

.643 

1.192 

6 

.105 

.995 

.105 

51 

.777 

.629 

1.235 

7 

.122 

.993 

.123 

52 

.788 

.616 

1.280 

8 

.139 

.990 

.141 

53 

.799 

.602 

1.327 

9 

.156 

.988 

.158 

54 

.809 

.58.8 

1.376 

10 

.174 

.985 

.176 

55 

.819 

.574 

1.428 

11 

.191 

.982 

.194 

56 

.829 

.559 

1.483 

12 

.208 

.978 

.213 

57 

.839 

.545 

1.540 

13 

.225 

.974 

.231 

58 

.848 

.530 

1.600 

14 

.242 

.970 

.249 

59 

.857 

.515 

1.664 

15 

-.259 

.966 

.268 

60 

.866 

.500 

1.732 

16 

.276 

.961 

.287 

61 

.875 

.485 

1.804 

17 

.292 

.956 

.306 

62 

.883 

.469 

1.881 

18 

.309 

.951 

.325 

63 

.891 

.454 

1.963 

19 

.326 

.946 

.344 

.64 

.899 

.438 

2.050 

20 

.342 

.940 

.364 

65 

.906 

.423 

2.144 

21 

.358 

.934 

.384 

66 

.914 

.407 

2.246 

22 

.375 

.927 

.404 

67 

.921 

.391 

2.356 

23 

.391 

.921 

.424 

68 

.927 

.375 

2.475 

24 

.407 

.914 

.445 

69 

.934 

.358 

2.605 

25 

.423 

.906 

.466 

70 

.940 

.342 

2.747 

26 

.438 

.899 

.488 

71 

.946 

.326 

2.904 

27    . 

.454 

.891 

.510 

72 

.951 

.309 

3.078 

28 

.469 

.883 

.532 

73 

.956 

.292 

3.271 

29 

.485 

.875 

.554 

74 

.961 

.276 

3.487 

30 

.500 

.866 

.577 

75 

.966 

.259 

3.732 

31 

.515 

.857 

.601 

76 

.970 

.242 

4.011 

32 

.530 

.848 

.625 

77 

.974 

.225 

4.331 

33 

.545 

.839 

.649 

78 

.978 

.208 

4.705 

34 

.559 

.829 

.675 

79 

.982 

.191 

5.145 

35 

.574 

.819 

.700 

80 

.985 

.174 

5.671 

36 

.588 

.809 

.727 

81 

.988   - 

.156 

6.314 

37 

.602 

.799 

.754 

82 

.990 

.139 

7.115 

38 

.616 

.788 

.781 

83 

.993 

.122 

8.144 

39 

.629 

.777 

.810 

84 

.995 

.105 

9.514 

40 

.643 

.766 

.839 

85 

.996 

.087 

11.430 

41 

.656 

.755 

.869 

86 

.998 

.070 

14.301 

42 

.669 

.743 

.900 

87 

.999 

.052 

19.081 

43 

.682 

.731 

.933 

88 

.999 

.035 

28.636 

44 

.695 

.719 

.966 

89 

.999 

.017 

57.290 

45 

.707 

.707 

1.000 

1 

UNITS   OF   MEASURE  183 

UNITS   OF   MEASURE 
234.     Units  of  Length 

ENGLISH 
12  inches  (in.)  =  1  foot  (ft.) 

3  feet  =  1  yard  (yd.) 
5J  yards  =  1  rod  (rd.) 
320  rods  or  5280  ft.  =  1  mile  (mi.)  . 

METRIC 

10  centimeters  (cm.)  =  1  decimeter  (dm.) 
10  decimeters  =1  meter  (m.) 
1000  meters  =  1  kilometer  (km.) 

1  meter  =  39. 37  in. 
1  kilometer  =  .  62  of  a  mile 

1  foot  =  30.48  centimeters 
1  mile  =1.6093  kilometers 

235.  Units  of  Surface 

ENGLISH 
144  square  inches  (sq.  in.)  =  1  square  foot  (sq.  ft.) 

9  square  feet  =  1  square  yard  (sq.  yd.) 
3034  square  yards  =  1  square  rod  (sq.  rd.) 

160  square  rods  =  1  acre  (A.) 
4840  square  yards  =  1  acre  (A.) 

640  acres  =  1  square  mile  (sq.  mi.) 

METRIC 

ICO  square  centimeters  =  1  square  decimeter 
100  square  decimeters  =  1  square  meter 

236.  Units  of  Volume 

ENGLISH  METRIC 

1728  cu.  in.  =  1  cu.  ft.  1000  cu.  mm.  =  1  cu.  cm. 

27  cu.  ft.  =  1  cu.  yd.  1000  cu.  cm.  =  1  cu.  dm. 

128  cu.  ft.  =  1  cord  (of  wood)  1000  cu.  dm.  =  1  cu.  m. 


184  SOLID   GEOMETRY 

237.  Units  of  Capacity 

ENGLISH 
Dry  Measure  Liquid  Measure 

2  pints  (pt.)  =  1  quart  (qt.)        4  gills  (gi.)  =  1  pint  (pt.) 
8  quarts  =  1  peck  (pk.)  2  pints  =  1  quart  (qt.) 

4  pecks  =1  bushel  (bu.)  4  quarts  =1  gallon  (gal.) 

1  bushel  =  2150. 42  cu.  in.         1  gallon  =  231  cu.  in.    . 

1   cu.    ft.    of    water    weighs 
62.4  Ibs. 

METRIC 
1  liter  =  1  cu.  dm. 

=  1000  cu.  cm. 
=  61.02  cu.  in. 
==  1 . 0567  liquid  quarts 
1  quart  =  9463  liters 

=  946.3  cu.  cm. 

238.  Units  of  Weight 

ENGLISH 

.  A  voirdupois  Weight  Troy  Weight 

16  ounces   (oz.)  =  l   pound  24  grains  (gr.)  =  l  penny- 

(lb,)  weight  (pwt.  or  dwt.) 

100    pounds  =1    hundred-  20  pwt.  =  1  Troy  ounce 

weight  (cwt.)  12    Troy   ounces  =1   Troy 

2000  pounds  =1  ton  (T.)  pound 

2240  pounds  =  1  long  ton  7000  grains  =1  avoirdupois  Ib. 

5760  grains  =  1  Troy  Ib. 

METRIC 
1  gram  =  weight  of  1  cu.  cm. 

of  water  at  39.2°  F. 
1000   grams  =  1   kilogram 

(kg.) 

1kg.  =2. 2046  Ibs. 
1  Ib.   =.45359  kg. 
=  453. 59  g. 


OUTLINE   OF  PLANE   GEOMETRY 

INTRODUCTORY  AND   GENERAL 

When  a  ray  starts  from  a  point  in  a  straight  line  and  forms  two 
congruent  angles,  the  angles  are  called  right  angles,  and  the  ray  is  said 
to  be  perpendicular  to  the  line. 

Two  angles  are  called  complementary  if  their  sum  is  an  angle  of  90°. 
Each  of  two  'Complementary  angles  is  called  the  complement  of  the 
other. 

Two  angles  are  said  to  be  supplementary  if  their  sum  is  an  angle  of 
180°.  Each  of  two  supplementary  angles  is  called  the  supplement 
of  the  other. 

As.  1.  Two  different  straight  lines  can  intersect  in  only  one  point, 
or  two  intersecting  straight  lines  locate  a  point. 

As.  2.     A  segment  can  have  only  one  mid-point. 

As.  3.  Only  one  segment  can  be  drawn  between  two  points,  or  a 
segment  is  located  definitely  if  its  extremities  are  given. 

As.  4.  Only  one  ray  can  be  drawn  having  a  given  origin  and  passing 
through  a  second  given  point. 

As.  5.     Only  one  ray  can  be  drawn  bisecting  a  given  angle. 

As.  6.     Only  one  straigh  t  line  can  pass  through  two  given  points. 

As.  7.  Only  one  perpendicular  can  be  drawn  to  a  line  from  a  given 
point  in  the  line. 

As.  8.  Only  one  perpendicular  can  be  drawn  to  a  line  from  a  given 
point  not  in  the  line. 

As.  9.     Circles  with  equal  radii  are  congruent. 

As.  10.  Congruent  circles  have  equal  radii. 

As.  11.  All  straight  angles  are  equal. 

As.  12.  All  right  angles  are  equal. 
•  As.  13.  Complements  of  equal  angles  are  equal. 

As.  14.  Supplements  of  equal  angles  are  equal. 

As.  15.  Vertical  angles  are  equal. 

As.  16.  If  a  ray  starts  from  a  point  in  a  straight  line,  the  sum  of  the 
two  adjacent  angles  formed  on  one  side  of  the  line  is  180°,  or  a  straight 
angle. 

As.  17.  The  sum  of  the  adjacent  angles  on  one  side  of  a  straight 
line  formed  by  any  number  of  rays  having  a  common  origin  on  the 
line  is  180°,  or  a  straight  angle. 

13  185 


186  SOLID   GEOMETRY 

As.  18.  The  sum  of  the  adjacent  angles  formed  by  a  number  of  rays 
from  the  same  origin  is  360°,  or  a  perigon. 

As.  19.  If  two  supplementary  angles  are  adjacent,  their  exterior 
sides  are  collinear. 

As.  20.  If  equal  segments  (or  angles)  are  added  to  equal  segments 
(or  angles),  the  results  are  equal  segments  (or  angles). 

As.  21.  If  equal  segments  (or  angles)  are  subtracted  from  equal 
segments  (or  angles),  the  results  are  equal  segments  (or  angles). 

As.  22.  If  equal  segments  (or  angles)  are  multiplied  by  the  same 
number,  the  results  are  equal  segments  (or  angles). 

As.  23.  If  equal  segments  (or  angles)  are  divided  by  the  same 
number,  the  results  are  equal  Segments  (or  angles). 

As.  24.  Segments  (or  angles)  that  are  equal  to  the  same  segment 
(or  angle)  are  equal. 

As.  25.  Equal  segments  (or  angles)  may  be  substituted  for  equal 
segments  (or  angles). 

CONGRUENT  TRIANGLES 

Any  two  figures  that  can  be  made  to  coincide  are  called  congruent 
figures. 

As.  26.  Any  figure  can  be  moved  about  in  space  without  changing 
either  its  size  or  its  shape. 

As.  27.  Figures  congruent  to  the  same  figure  are  congruent  to  each 
other. 

THEOREM  1.  If  two  sides  and  the  included  angle  of  one  triangle  are 
equal  to  two  sides  and  the  included  angle  of  another  triangle,  the  tri- 
angles are  congruent  in  all  corresponding  parts  and  are  called  congruent 
triangles. 

THEOREM  2.  If  two  angles  and  the  included  side  of  one  triangle  are 
equal  to  two  angles  and  the  included  side  of  another  triangle,  the  tri- 
angles are  congruent  in  all  corresponding  parts  and  are  called  congruent 
triangles. 

THEOREM  4.  If  three  sides  of  one  triangle  are  equal  to  three  sides 
of  another  triangle,  the  triangles  are  congruent. 

THEOREM  22.  Two  right  triangles  are  congruent  if  the  hypotenuse 
and  an  acute  angle  of  one  are  equal  to  the  hypotenuse  and  an  acute 
angle  of  the  other. 

THEOREM  23.  Two  right  triangles  are  congruent  if  the  hypotenuse 
and  a  side  of  one  are  equal  t6  the  hypotenuse  and  a  side  of  the  other. 

COR.  If  a  perpendicular  is  erected  to  a  straight  line,  equal  segments 
drawn  from  the  same  point  in  the  perpendicular  cut  off  equal  distances 
from  the  foot  of  the  perpendicular. 


OUTLINE   OF  PLANE   GEOMETRY  187 

THEOREM  5.  If  a  perpendicular  be  erected  to  a  straight  line, 
oblique  segments  drawn  from  the  same  point  in  the  perpendicular  cut- 
ting the  straight  line  at  equal  distances  from  the  foot  of  the  perpen- 
dicular are  equal. 

PARALLELS   AND   ANGLES 

As.  ;•>().  ( )nly  one  line  can  be  drawn  through  a  given  point  parallel 
to  a  given  line. 

THEOREM  9.  If  two  straight  lines  in  the  same  plane  are  cut  by  a 
third  straight  line  so  that  the  alternate  interior  angles  are  equal,  the 
two  straight  lines  arc  parallel. 

THEOREM  10.  If  two  straight  lines  in  the  same  plane  are  cut  by  a 
third  straight  line  so  that  one  pair  of  corresponding  angles  are  equal, 
the  two  straight  lines  are  parallel. 

THEOREM  11.  If  two  straight  lines  in  the  same  plane  are  cut  by  a 
third  straight  line  so  that  the  interior  angles  on  the  same  side  of  the 
transversal  are  supplements,  the  two  straight  lines  are  parallel. 

THKMKKM  12.  Two  straight  lines  in  the  same  plane  perpendicular 
to  the  same  straight  line  are  parallel. 

THKOKKM  n.  Two  lines  parallel  to  a  third  line  are  parallel  to  each 
other. 

THKOKKM  II.  If  two  parallel  lines  are  cut  by  a  third  straight  line, 
the  alternate  interior  angles  are  equal. 

THEOREM  15.  If  two  parallel  lines  are  cut  by  a  third  straight  line, 
the  corresponding  angles  are  equal. 

THEOREM  16.  If  two  parallel  lines  are  cut  by  a  third  straight  line, 
the  interior  angles  on  the  same  side  of  the  transversal  are  supplements 
of  each  other. 

THEOREM  17.  A  line  which  is  perpendicular  to  one  of  two  par- 
allels is  perpendicular  to  the  other. 

ANGLE   SUMS 

THEOREM  18.  The  sum  of  the  interior  angles  of  a  triangle  is  two 
right  angles. 

COR.  I.     Each  angle  of  an  equilateral  triangle  is  60°. 

COR.  II.  If  two  angles  of  one  triangle  are  equal  respectively  to 
two  angles  of  a  second  triangle,  the  third  angles  are  equal.  • 

COR.  III.  The  acute  angles  of  a  right  triangle  are  complements 
of  each  other. 

THEOREM  19.  The  exterior  angle  of  a  triangle  is  equal  to  the  sum  of 
the  two  non-adjacent  interior  angles. 

THEOREM  20.  The  sum  of  the  interior  angles  of  a  polygon  of  n  sides 
is  2(n—  2)  right  angles. 

13A 


188  .  SOLID   GEOMETRY 

THEOREM  21.  The  sum  of  the  exterior  angles  of  a  polygon  of  n  sides 
is  four  right  angles. 

ISOSCELES   TRIANGLES 

A  triangle  that  has  at  least  two  sides  equal  is  called  an  isosceles 
triangle. 

THEOREM  3.  The  angles  opposite  the  equal  sides  of  an  isosceles 
triangle  are  equal. 

COR.  An  equilateral  triangle  has  all  of  its  angles  equal;  that  is, 
it  is  equiangular. 

THEOREM  24.  If  two  angles  of  a  triangle  are  equal,  the  triangle 
is  isosceles. 

THEOREM  6.  The  bisector  of  the  vertex  angle  of  an  isosceles  triangle 
is  the  perpendicular  bisector  of  the  base. 

THEOREM  7.  The  segment  which  joins  the  vertex  of  an  isosceles 
triangle  with  the  mid-point  of  the  base  bisects  the  vertex  angle  and 
is  perpendicular  to  the  base 

THEOREM  25.  A  segment  from  the  vertex  of  an  isosceles  triangle 
perpendicular  to  the  base  bisects  the  base  and  the  vertex  angle. 

THEOREM  26.  The  bisector  of  the  vertex  angle  of  an  isosceles  tri- 
angle is  an  axis  of  symmetry  of  the  triangle. 

SYMMETRY 

A  figure  is  said  to  be  symmetric  with  respect  to  a  line  as  an  axis  if 

one  part  coincides  with  the  remainder  when  it  is  folded  on  that  line  as 
an  axis.  Two  figures  are  said  to  be  symmetric  with  respect  to  a  line  as 
an  axis  if  one  figure  coincides  with  the  other  when  the  plane  in  which 
it  lies  is  folded  on  that  line  as  an  axis. 

Such  a  figure  or  such  figures  are  said  to  have  axial  symmetry. 

A  figure  is  said  to  be  symmetric  with  respect  to  a  point  as  a  center 
if  one  part  of  the  figure  coincides  with  the  remainder  when  it  is  rotated 
through  an  angle  of  180°  about  the  point  as  a  center.  Two  figures  are 
said  to  be  symmetric  with  respect  to  a  point  as  a  center  if  one  figure 
coincides  with  the  other  when  it  is  rotated  through  an  angle  of  180° 
about  the  point  as  a  center. 

Such  a  figure  or  such  figures  are  said  to  have  central  symmetry. 

THEOREM  27.  Two  polygons  are  symmetric  with  respect  to  an  axis 
if  the  vertices  of  one  are  symmetric  to  the  corresponding  vertices  of 
the  other. 

THEOREM  28.  Two  polygons  are  symmetric  with  respect  to  a  center 
if  the  vertices  of  one  are  symmetric  to  the  corresponding  vertices  of 
the  other. 


OUTLINE   OF   PLANE   GEOMETRY  189 

THEOREM  29.  Any  figure  that  has  two  axes  of  symmetry  at  right 
angles  to  each  other  has  the  intersection  of  the  axes  as  a  center  of 
symmetry. 

QUADRILATERALS 

A  quadrilateral  with  each  side  parallel  to  its  opposite  is  called  a 
parallelogram. 

The  perpendicular  distance  between  the  bases  of  a  parallelogram  is 
called  the  altitude  of  the  parallelogram. 

A  quadrilateral  with  but  one  pair  of  parallel  sides  is  called  a  trapezoid. 

If  a  trapezoid  has  its  non-parallel  sides  equal,  it  is  called  an  isosceles 
trapezoid. 

A  |  >arallelogram  with  one  right  angle  is  called  a  rectangle. 

A  parallelogram  with  two  consecutive  sides  equal  is  called  a  rhombus. 

A  rec tangle  with  two  consecutive  sides  equal  is  called  a  square. 

The  segment  joining  the  mid-points  of  two  opposite  sides  of  a  quadri- 
lateral is  called  a  median  of  the  quadrilateral. 

THEOREM  30.  Each  diagonal  of  a  parallelogram  divides  it  into  two 
congruent  triangles. 

THEOREM  31.     The  opposite  sides  of  a  parallelogram  are  equal. 

THEOREM  32.     The  opposite  angles  of  a  parallelogram  are  equal. 

THEOREM  :J3.     The  diagonals  of  a  parallelogram  bisect  each  other. 

THEOREM  34.  The  intersection  of  the  diagonals  of  a  parallelogram 
is  the  center  of  symmetry  of  the  parallelogram. 

THEOREM  3").  Two  parallelograms  are  congruent  if  two  sides  and 
the  included  angle  of  one  are  equal  to  two  sides  and  the  included  angle 
of  the  other. 

THEOREM  36.  If  a  quadrilateral  has  one  side  equal  and  parallel  to 
its  opposite,  it  is  a  parallelogram. 

THEOREM  37.  If  a  quadrilateral  has  each  side  equal  to  its  opposite, 
it  is  a  parallelogram. 

THEOREM  38.  If  the  diagonals  of  a  quadrilateral  bisect  each  other, 
the  quadrilateral  is  a  parallelogram. 

THEOREM  39.  Segments  of  parallels  intercepted  between  parallel 
lines  are  equal. 

THEOREM  40.  Segments  of  perpendiculars  intercepted  between 
parallel  lines  are  equal. 

THEOREM  41.  The  diagonals  of  a  kite  are  perpendicular  to  each 
other,  and  the  one  which  is  an  axis  of  symmetry  bisects  the  other. 

THEOREM  *42.     All  the  angles  of  a  rectangle  are  right  angles. 

THEOREM  43.     All  the  sides  of  a  rhombus  are  equal. 

THEOREM  44.  The  diagonals  of  a  rhombus  are  perpendicular  to 
each  other  and  bisect  the  angles  through  which  they  pass. 


190  SOLI©   GEOMETRY 

PARALLELS  AND   TRANSVERSALS 

THEOREM  45.  If  a  series  of  parallels  cuts  off  equal  segments  on  one 
transversal,  it  cuts  off  equal  segments  on  all  transversals. 

THEOREM  46.  A  segment  parallel  to  the  base  of  a  triangle  and  bisect- 
ing one  side  is  equal  to  half  the  base. 

THEOREM  47.  A  segment  parallel  to  the  base  of  a  triangle  and  bisect- 
ing one  side  bisects  the  other  side  also. 

THEOREM  48.  A  segment  bisecting  two  sides  of  a  triangle  is  parallel 
to  the  third  side. 

The  segment  joining  any  vertex  of  a  triangle  with  the  mid-point  of 
the  opposite  side  is  called  a  median  of  the  triangle. 

THEOREM  50.  The  median  from  the  vertex  of  the  right  angle  of  a 
right  triangle  to  the  hypotenuse  is  one-half  the  hypotenuse. 

THEOREM  51.  The  segment  joining  the  mid-points  of  the  non- 
parallel  sides  of  a  trapezoid  is  parallel  to  the  bases. 

THEOREM  52.  The  segment  joining  the  mid-points  of  the  non- 
parallel  sides  of  a  trapezoid  is  equal  to  one-half  the  sum  of  the  bases. 

INEQUALITIES 

As.  28.  If  one  angle  or  segment  is  greater  than  a  second  and  the 
second  is  equal  to  or  greater  than  a  third,  then  the  first  is  greater  than 
the  third. 

As.  29.     The  whole  is  greater  than  any  of  its  parts. 

As.  31.  If  equal  segments  (or  angles)  are  added  to  unequal  seg- 
ments (or  angles),  the  resulting  segments  (or  angles)  are  unequal  in 
the  same  order. 

As.  32.  If  equal  segments  (or  angles)  are  subtracted  from  unequal 
segments  (or  angles),  the  resulting  segments  (or  angles)  are  unequal 
in  the  same  order. 

As.  33.  If  unequal  segments  (or  angles)  are  added  to  unequal  seg- 
ments (or  angles),  the  greater  to  the  greater  and  the  lesser  to  the 
lesser,  the  resulting  segments  (or  angles)  are  unequal  in  the  same  order. 

As.  34.  If  unequal  segments  (or  angles)  are  subtracted  from  equal 
segments  (or  angles),  the  resulting  segments  (or  angles)  are  unequal  in 
the  opposite  order. 

As.  35.  If  unequal  segments  (or  angles)  are  multiplied  by  the  same 
number,  the  resulting  segments  (or  angles)  are  unequal  in  the  same 
order. 

As.  36.  If  unequal  segments  (or  angles)  are  divided  by  the  same 
number,  the  resulting  segments  (or  angles)  are  unequal  in  the  same 
order. 


OUTLINE   OF   PLANE   GEOMETRY  191 

As.  37.  The  sum  of  two  sides  of  a  triangle  is  greater  than  the 
third. 

As.  38.  The  difference  between  two  sides  of  a  triangle  is  less  than 
the  third  side. 

THEOREM  8.  An  exterior  angle  of  a  triangle  is  greater  than  either 
of  the  non-adjacent  interior  angles. 

THEOREM  53.     If  from  a  point  within  a  triangle  segments  are  drawn  • 
to  the  extremities  of  one  side,  their  sum  is  less  than  the  sum  of  the 
other  two  sides  of  the  triangle. 

THEOREM  54.  If  one  angle  of  a  triangle  is  greater  than  a  second, 
the  side  opposite  the  first  angle  is  greater  than  the  side  opposite  the 
second  angle. 

THEOREM  55.  If  one  side  of  a  triangle  is  greater  than  a  second,  the 
angle  opposite  the  greater  side  is  greater  than  the  angle  opposite  the 
lesser  side. 

THEOREM  56.  The  perpendicular  is  the  shortest  segment  from  a 
point  to  a  straight  line-. 

THEOREM  57.  If  from  a  point  in  a  perpendicular  to  a  straight  line 
two  oblique  segments  are  drawn  cutting  the  straight  line  at  unequal 
distances  from  the  foot  of  the  perpendicular,  the  more  remote  is  the 
greater. 

THEOREM  58.  If  from  a  point  in  a  perpendicular  to  a  straight  line 
two  unequal  oblique  segments  are  drawn,  the  greater  cuts  the  straight 
line  at  the  greater  distance  from  the  foot  of  the  perpendicular. 

THEOREM  59.  If  two  triangles  have  two  sides  of  one  equal  to  two 
sides  of  the  other,  but  the  included  angle  of  one  greater  than  the 
included  angle  of  the  other,  the  third  side  of  the  first  is  greater  than  the 
third  side  of  the  second. 

THEOREM  60.  Jf  two  triangles  have  two  sides  of  one  equal  to  two 
sides  of  the  other,  but  the  third  side  of  one  greater  than  the  third  side 
of  the  other,  the  angle  opposite  the  third  side  of  the  first  is  greater  than 
the  angle  opposite  the  third  side  of  the  second. 

CIRCLES 

A  closed  curved  line  every  point  of  which  is  equally  distant  from  a 
given  point  in  the  same  plane  is  called  a  circle.  The  given  point  is 
ealled  the  center  of  the  circle. 

As.  39.     The  diameter  of  a  circle  is  twice  its  radius. 

As.  K).  A  circle  is  located  definitely  if  its  center  and  its  radius  are 
known. 

As.  41.  If  a  line  passes  through  a  point  within  a  circle,  the  line  and 
the  circle  intersect  in  two  and  only  two  points. 


192  SOLID   GEOMETRY 

As.  42.     Every  diameter  bisects  the  circle, 

As.  43.  A  circle  is  symmetric  with  respect  to  any  diameter  as  an 
axis  and  with  respect  to  its  center  as  a  center. 

As.  44.  Between  the  same  two  points  on  a  circle  there  is  one  and 
only  one  minor  arc  of  the  circle,  provided  these  points  are  not  the  ends 
of  a  diameter. 

•    As.  45.     A  segment  joining  a  point  within  a  circle  and  the  center  is 
shorter  than  the  radius. 

As.  46.  If  a  segment  that  has  one  end  at  the  center  of  a  circle  is 
shorter  than  the  radius,  it  lies  wholly  within  the  circle. 

As.  47.  A  segment  joining  a  point  without  a  circle  and  the  center 
is  longer  than  the  radius. 

As.  48.  If  a  segment  that  has  one  end  at  the  center  of  a  circle  is 
longer  than  the  radius,  it  extends  without  the  circle  and  cuts  the  circle 
but  once. 

As.  49.  In  the  same  circle  or  in  congruent  circles  equal  central 
angles  intercept  equal  minor  arcs. 

As.  50.  In  the  same  circle  or  in  congruent  circles  equal  minor  arcs 
intercept  equal  central  angles. 

As.  54.  In  the  same  circle,  or  in  congruent  circles,  if  two  central 
angles  are  unequal,  the  minor  arc  subtended  by  the  greater  angle  is 
greater  than  the  minor  arc  subtended  by  the  lesser  angle. 

As.  55.  In  the  same  circle  or  in  congruent  circles,  if  two  minor  arcs 
are  unequal,  the  angle  subtended  by  the  greater  arc  is  greater  than  the 
angle  subtended  by  the  lesser  arc. 

CIRCLES  AND  RELATED  LINES 

A  line  that  touches  a  circle  at  one  point  but  does  not  cut  it  is  called 
a  tangent  to  the  circle.  This  definition  is  the  fundamental  test  for 
tangents. 

The  point  at  which  the  tangept  touches  the  circle  is  called  the 
point  of  contact  or  the  point  of  tangency  of  the  tangent. 

THEOREM  61.     In  the  same  circle  or  in  congruent  circles 

A.  Equal  chords  intercept  equal  central  angles. 

B.  Equal  central  angles  intercept  equal  chords. 
THEOREM  62.     In  the  same  circle  or  in  congruent  circles 

A.  Equal  chords  have  equal  minor  arcs. 

B.  Equal  minor  arcs  have  equal  chords. 

THEOREM  63.  A  radius  perpendicular  to  a  chord  bisects  the  chord 
and  its  arc. 

THEOREM  64.  The  perpendicular  bisector  of  a  chord  passes  through 
the  center  of  the  circle. 


OUTLINE   OF   PLANE   GEOMETRY  193 

THEOREM  65.  One  and  only  one  circle  can  be  drawn  through  three 
non-collinear  points. 

THEOREM  66.  If  in  the  same  circle  or  in  congruent  circles  perpen- 
diculars from  the  center  to  two  chords  are  equal,  the  chords  are  equal. 

THEOREM  67.  In  the  same  circle  or  in  congruent  circles  perpendic- 
ulars from  the  center  to  two  equal  chords  are  equal. 

THEOREM  68.  A  line  which  is  perpendicular  to  a  radius  at  its  outer 
extremity  is  a  tangent  to  the  circle. 

THEOREM  69.  A  tangent  to  a  circle  is  perpendicular  to  the  radius 
drawn  to  the  point  of  contact. 

THEOREM  70.  If  two  tangents  meet  at  a  point  without  a  circle,  the 
distances  from  the  intersection  to  the  points  of  tangency  are  equal. 

THEOREM  71.  A  perpendicular  to  a  tangent  at  the  point  of  contact 
passes  through  the  center  of  the  circle. 

MEASUREMENT   OF  ANGLES 

As.  53.  The  measure  of  a  central  angle  and  its  intercepted  arc  are 
expressed  by  the  same  number,  or  a  central  angle  is  measured  by  its 
intercepted  arc. 

An  angle  is  said  to  be  inscribed  in  a  circle  if  its  vertex  is  on  the 
circle  and  its  sides  are  chords  of  the  circle. 

The  arc  cut  off  between  the  sides  of  an  inscribed  angle  is  called  its 
intercepted  arc. 

THEOREM  77.  An  inscribed  angle  is  measured  by  one-half  its  inter- 
cepted arc. 

COR.  I.  Inscribed  angles  measured  by  the  same  or  by  equal  arcs 
are  equal,  and,  conversely,  arcs  that  measure  equal  inscribed  angles 
are  equal. 

COR.  II.     An  angle  inscribed  in  a  semicircle  is  a  right  angle. 

COR.  III.  Inscribed  angles  are  supplementary  if  the  sum  of  their 
intercepted  arcs  is  360°. 

THEOREM  78.  An  angle  formed  by  two  chords  intersecting  within  a 
circle  is  measured  by  one-half  the  sum  of  the  intercepted  arcs. 

THEOREM  79.  An  angle  formed  by  two  secants  intersecting  without 
a  circle  is  measured  by  one-half  the  difference  of  the  intercepted  arcs. 

THEOREM  80.     Parallel  chords  intercept  equal  arcs  on  a  circle. 

THEOREM  81 .  An  angle  formed  by  a  tangent  and  a  chord  is  measured 
by  one-half  its  intercepted  arc. 

THEOREM  82.  If  a  chord  and  a  tangent  are  parallel,  they  cut  off  equal 
arcs. 

THEOREM  83.  An  angle  formed  by  a  secant  and  a  tangent  is  meas- 
ured by  one-half  the  difference  of  the  intercepted  arcs. 


194  SOLID   GEOMETRY 

THEOREM  84.  An  angle  formed  by  two  tangents  is  measured  by 
one-half  the  difference  of  the  intercepted  arcs. 

TWO    CIRCLES 

As.  51.  The  line  of  centers  of  two  circles  is  an  axis  of  symmetry 
of  the  two  circles.  * 

THEOREM  72.  If  two  circles  intersect  in  one  point  not  on  the  line 
of  centers,  they  intersect  in  two  points. 

COR.  If  two  circles  intersect,  the  points  of  intersection  are  sym- 
metric points. 

As.  52.     Two  circles  cannot  intersect  at  more  than  two  points. 

THEOREM  73.  If  any  two  circles  intersect,  the  line  of  centers  is  the 
perpendicular  bisector  of  the  common  chord. 

THEOREM  74.  If  two  congruent  circles  intersect,  the  common  chord 
is  an  axis  of  symmetry  of  the  figure. 

COR.  If  t.wo  congruent  circles  intersect,  the  segment  joining  the 
centers  and  the  common  chord  are  perpendicular  bisectors  of  each 
other. 

Two  circles  are  said  to  be  tangent  if  they  have  but  one  common 
point.  They  may  be  tangent  internally  or  tangent  externally. 

THEOREM  75.  If  two  circles  meet  at  a  point  on  their  line  of  centers, 
the  circles  are  tangent. 

COR.  I.  If  the  segment  joining  the  centers  of  two  circles  is  equal  to 
the  sum  of  the  radii,  the  circles  are  tangent  externally. 

COR.  II.  If  the  segment  joining  the  centers  of  two  circles  is  equal 
to  the  difference  between  the  radii,  the  circles  are  tangent  internally. 

THEOREM  76.  If  two  circles  are  tangent,  the  point  of  contact  is  on 
the  line  of  centers. 

LOCI 

A  point  which  moves  so  as  to  fulfill  some  given  requirement  is  called 
a  variable  point. 

The  path  of  a  point  which  moves  so  as  to  fulfill  some  given  require- 
ment is  called  a  locus. 

A  line  or  group  of  lines  is  called  a  locus  if  they  contain  all  points 
which  fulfill  some  given  requirement  and  contain  no  other  points. 

THEOREM  85.  The  bisector  of  an  angle  is  the  locus  of  points  equally 
distant  from  the  sides  of  the  angle. 

THEOREM  86.  The  perpendicular  bisector  of  a  segment  is  the  locus 
of  a  point  equally  distant  from  the  ends  of  the  segment! 

COR.  If  two  points  are  each  equally  distant  from  the  extremities 
of  a  segment,  the  line  passing  through  these  points  is  the  perpendicular 
bisector  of  the  segment. 


OUTLINE   OF   PLANE   GEOMETRY  195 

THEOREM  87.  The  perpendicular  bisectors  of  the  sides  of  a  triangle 
are  concurrent  at  a  point  which  is  equally  distant  from  the  vertices. 

A  perpendicular  from  any  vertex  of  a  triangle  to  the  opposite  side  is 
called  an  altitude  of  the  triangle. 

THEOREM  88.     The  altitudes  of  a  triangle  are  concurrent. 

THEOREM  89.  The  bisectors  of  the  angles  of  a  triangle  are  concurrent 
at  a  point  equally  distant  from  the  sides  of  the  triangle. 

THEOREM  49.  The  medians  of  a  triangle  are  concurrent  in  a  point 
that  is  two-thirds  the  distance  from  each  vertex  to  the  mid-point  of 
the  opposite  side. 

Each  triangle  has  four  sets  of  concurrent  lines.  The  intersection  of 
each  set  has  a  special  name  as  shown  below. 

I.  The  medians .  .  : centroid  or  center  of  gravity 

II.  Perpendicular  bisectors  of  the  sides circumcenter 

III.  The  altitudes orthocenter 

IV.  Bisectors  of  the  angles incenter 

RATIOS 

As.  56.  Multiplying  or  dividing  both  terms  of  a  ratio  by  the  same 
number  does  not  change  the  value  of  the  ratio. 

As.  o7.     Ratios  equal  to  the  same  ratio  are  equal. 

As.  58.     Equal  ratios  may  be  substituted  for  equal  ratios. 

THKOREM  90.  If  four  numbers  are  in  proportion,  the  product  of 
the  means  is  equal  to  the  product  of  the  extremes. 

THEOREM  91.  If  the  product  of  two  numbers  equals  the  product 
of  two  other  numbers,  either  pair  of  factors  may  be  made  the  extremes 
and  the  other  pair  the  means  of  a  proportion. 

THEOREM  92.  If  three  terms  of  one  proportion  are  equal  respectively 
to  three  corresponding  terms  of  another  proportion,  the  fourth  terms 
are  equal. 

THEOREM  93.  If  four  numbers  are  in  proportion,  the  first  is  to  the 
third  as  the  second  is  to  the  fourth;  that  is,  they  are  in  proportion  by 
mean  alternation. 

THEOREM  94.  If  four  numbers  are  in  proportion,  the  fourth  is  to 
the  second  as  the  third  is  to  the  first;  that  is,  they  are  in  proportion  by 
extreme  alternation. 

THEOREM  95.  If  four  numbers  are  in  proportion,  the  second  is  to 
the  first  as  the  fourth  is  to  the  third;  that  is,  they  are  in  proportion  by 
in  version. 

THEOREM  9(i.  If  four  numbers  are  in  proportion,  the  first  plus  the 
second  is  to  the  second  as  the  third  plus  the -fourth  is  to  the  fourth; 
that  is,  they  are  in  proportion  by  addition.  This  is  sometimes  called 
proportion  by  composition. 


196  SOLID   GEOMETRY 

THEOREM  97.  If  four  numbers  are  in  proportion,  the  first  minus 
the  second  is  to  the  second  as  the  third  minus  the  fourth  is  to  the  fourth ; 
that  is,  they  are  in  proportion  by  subtraction.  This  is  sometimes  called 
proportion  by  division. 

THEOREM  124.  In  a  series  of  equal  ratios  the  sum  of  the  antecedents 
is  to  the  sum  of  the  consequents  as  any  antecedent  is  to  its  consequent. 

THEOREM  98.  If  three  parallels  cut  two  transversals,  the  segments 
on  one  transversal  have  the  same  ratio  as  the  corresponding  segments 
on  the  other  transversal. 

THEOREM  99.  If  a  line  is  parallel  to  the  base  of  a  triangle,  the  ratio 
of  the  segments  on  one  side  equals  the  ratio  of  the  corresponding  seg- 
ments on  the  other  side. 

COR.  If  a  line  is  parallel  to  the  base  of  a  triangle,  one  side  is  to  either 
of  its  segments  as  the  other  side  is  to  its  corresponding  segment. 

THEOREM  100.  If  a  line  divides  the  sides  of  a  triangle  so  that  one 
side  is  to  one  segment  as  a  second  side  is  to  its  corresponding  segment, 
the  line  is  parallel  to  the  third  side  of  the  triangle. 

COR.  If  a  line  divides  the  sides  of  a  triangle  so  that  the  ratio  of  the 
segments  on  one  side  is  equal  to  the  ratio  of  the  segments  on  the  other, 
the  line  is  parallel  to  the  third  side  of  the  triangle. 

THEOREM  101.  If  two  triangles  have  the  angles  of  one  respectively 
equal  to  the  angles  of  the  other,  the  corresponding  sides  have  equal 
ratios. 

THEOREM  102.     Two  mutually  equiangular  triangles  are  similar.. 

If  the  product  of  two  segments  equals  the  square  of  a  third  segment, 
the  last  segment  is  called  a  mean  proportional  between  the  other"  two. 
In  b2  =  cm,  b  is  a  mean  proportional  between  c  and  m. 

Our  fundamental  methods  for  proving  ratios  equal  are : 

1.  By  parallels  and  transversals. 

2.  By  similar  triangles. 

Before  either  of  these  methods  can  be  applied  it  is  often  necessary  to 
find  a  third  ratio  to  which  each  of  the  given  ratios  can  be  proved  equal. 

THEOREM  103.  If  two  chords  intersect  within  a  circle,  the  product 
of  the  segments  of  one  is  eqxial  to  the  product  of  the  segments  of  the 
other. 

THEOREM  104.  If  two  secants  intersect  without  a  circle,  the  product 
of  one  secant  and  its  external  segment  is  equal  to  the  product  of  the 
other  secant  and  its  external  segment. 

THEOREM  105,  If  a  secant  and  a  tangent  meet  without  a  circle, 
the  tangent  is  a  mean  proportional  between  the  whole  secant  and  its 
external  segment. 


OUTLINE   OF   PLANE   GEOMETRY  197 

THEOREM  106.  The  bisector  of  an  angle  of  a  triangle  divides  the 
opposite  side  internally  into  segments  that  have  the  same  ratio  as  the 
other  two  sides  of  the  triangle. 

THEOREM  107.  The  bisector  of  an  exterior  angle  of  a  triangle  divides 
the  opposite  side  externally  into  segments  that  have  the  same  ratio  as 
the  other  two  sides  of  the  triangle. 

THEOREM  108.  If  a  perpendicular  is  drawn  from  the  vertex  of  the 
right  angle  of  a  right  triangle  to  the  hypotenuse,  the  perpendicular  is 
a  mean  proportional  between  the  segments  of  the  hypotenuse. 

THEOREM  109.  If  a  perpendicular  is  drawn  from  the  vertex  of  the 
right  angle  of  a  right  triangle  to  the  hypotenuse,  either  leg  is  a  mean 
proportional  between  the  whole  hypotenuse  and  the  segment  adjacent 
to  that  leg. 

THEOREM  110.  The  sum  of  the  squares  of  the  legs  of  a  right  tri- 
angle is  equal  to  the  square  of  the  hypotenuse. 

THEOREM  111.  If  one  side  of  a  square  is  s,_its  diagonal  is  sV2  .  If 
the  diagonal  of  a  square  is  d,  the  side  is  %  d  V2  . 

THEOREM  112.  If  one  side  of  an  equilateral  triangle  is  s,  its  altitude 
is  Yz  sV3  .  If  the  altitude  is  a,  one  side  of  the  equilateral  triangle  is 


SIMILAR  FIGURES 

Similar  polygons  are  polygons  that  have 

1.  The  angles  of  one  equal  to  the  corresponding  angles  of  the  other, 
and 

2.  Corresponding  sides  proportional. 

The  ratio  of  similitude  of  two  similar  polygons  is  the  ratio  of  any  two 
corresponding  sides. 

If  the  segments  which  join  the  vertices  of  a  polygon  with  a  given  point 
are  divided  in  the  same  ratio  from  the  given  point  and  the  points  of 
division  joined  in  the  same  order  as  the  vertices  of  the  polygon,  the 
polygon  so  formed  ami  the  ^iven  polygon  are  radially  placed.  The 
point  may  be  without  the  polygon,  or  within  the  polygon.  The  radial 
point  is  called  the  center  of  similitude  of  the  polygons. 

THEOREM  102.     Two  mutually  equiangular  triangles  are  similar. 

THEOREM  119.  Two  triangles  are  similar  if  an  angle  of  one  is  equal 
to  an  angle  of  the  other  and  the  ratios  of  the  including  sides  are  equal. 

THEOREM  120.  Two  triangles  are  similar  if  the  corresponding  sides 
have  equal  ratios. 

THEOREM  121.  Two  polygons  are  similar  if  diagonals  drawn  from 
two  corresponding  vertices  divide  the  polygons  into  the  same  number 
of  triangles  similar  each  to  each  and  similarly  placed. 


198  SOLID   GEOMETRY 

THEOREM  122.  If  two  polygons  are  similar,  diagonals  drawn  from 
two  corresponding  vertices  divide  the  polygon  into  the  same  number 
of  triangles  similar  each  to  each  and  similarly  placed. 

THEOREM  123.  The  ratio  of  corresponding  altitudes  of  two  similar 
triangles  equals  the  ratio  of  the  bases. 

THEOREM  125.  The  ratio  of  the  perimeters  of  two  similar  triangles 
is  equal  to  the  ratio  of  similitude. 

THEOREM  126.  The  areas  of  two  similar  triangles  have  the  same 
ratio  as  the  squares  of  the  bases  or  the  squares  of  the  altitudes. 

THEOREM  127.  The  areas  of  two  similar  polygons  have  the  same 
ratio  as  the  squares  of  two  corresponding  sides. 

REGULAR  POLYGONS 

A  polygon  of  3  sides  is  called  a  triangle. 

A  polygon  of  4  sides  is  called  a  quadrilateral. 

A  polygon  of  5  sides  is  called  a  pentagon. 

A  polygon  of  6  sides  is  called  a  hexagon. 

A  polygon  of  7  sides  is  called  a  heptagon. 

A  polygon  of  8  sides  is  called  an  octagon. 

A  polygon  of  10  sides  is  called  a  decagon. 

A  polygon  of  12  sides  is  called  a  duodecagon. 

A  polygon  of  15  sides  is  called  a  pentadecagon. 

A  polygon  with  all  of  its  sides  and  alt  of  its  angles  equal  is  a  regular 
polygon. 

A  polygon  is  said  to  be  inscribed  in  a  circle  if  its  vertices  are  on  the 
circle  and  its  sides  are  chords  of  the  circle.  In  this  case  the  circle  is 
said  to  be  circumscribed  about  the  polygon.  A  polygon  is  said  to  be 
circumscribed  about  a  circle  if  its  sidss  are  tangant  to  the  circle.  In 
this  case  the  circle  is  said  to  be  inscribed  in  the  polygon. 

THEOREM  128.  If  a  circle  is  divided  into  n  equal  arcs,  the  chords 
joining  the  points  of  division  form  a  regular  polygon. 

THEOREM  129.  If  a  circle  is  divided  into  n  equal  arcs,  the  tangents 
drawn  fo  the  points  of  division  form  a  regular  polygon. 

To  construct  regular  4-,  8-,  or  16-sided  polygons,  construct  two 
perpendicular  diameters. 

To  construct  regular  3-,  6-,  or  12-sided  polygons,  construct  a  central 
angle  of  60°  by  means  of  an  equilateral  triangle. 

To  construct  regular  5-,  10-,  or  16-sided  polygons,  divide  the  radius 
of  the  circle  into  extreme  and  mean  ratio. 

THEOREM  130.  A  circle  can  be  circumscribed  about  any  regular 
polygon. 


OUTLINE   OF   PLANE   GEOMETRY  199 

COR.  The  radius  of  the  circumscribed  circle  of  a  regular  polygon 
bisects  the  angle  through  whose  vertex  it  passes. 

THEOREM  131.     A  circle  can  be  inscribed  in  any  regular  polygon. 

The  center  of  the  circumscribed  and  of  the  inscribed  circle  of  a 
regular  polygon  is  called  the  center  of  the  polygon. 

The  radius  of  the  circumscribed  circle  of  a  regular  polygon  is  called 
the  radius  of  the  polygon. 

The  radius  of  the  inscribed  circle  of  a  regular  polygon  is  called  the 
apothem  of  the  polygon. 

By  the  central  angle  of  a  regular  polygon  is  meant  the  angle  between 
two  consecutive  radii. 

COR.  I.  The  central  angle  of  a  regular  polygon  of  n  sides  is  l/n  of 
360°. 

COR.  II.  The  radius  of  a  regular  polygon  bisects  the  angle  between 
two  consecutive  apothems,  and  the  apothem  bisects  the  angle  between 
two  consecutive  radii. 

COR.  III.  The  radius  of  a  regular  polygon  bisects  the  arc  between 
the  points  of  contact  of  the  inscribed  circle. 

THEOREM    132.  •  Each   angle   of   a   regular   polygon  of   n  sides  is 

2w-4    4 

rt.   A. 

n 

THEOREM  133.  The  area  of  a  regular  polygon  is  one-half  the  product 
of  the  perimeter  and  the  apothem. 

THEOREM  134.  Two  regular  polygons  of  the  same  number  of  sides 
are  similar. 

THEOREM  135.  If  two  regular  polygons  have  the  same  number  of 
sides,  the  ratio  of  the  perimeters  is  equal  to  the  ratio  of  the  radii  or  of 
the  apothems. 

COR.  The  ratio  of  the  perimeter  to  the  diameter  of  the  inscribed 
or  of  the  circumscribed  circle  is  the  same  for  all  regular  polygons  of  the 
same  number  of  sides. 

THEOREM  136.  If  two  regular  polygons  have  the  same  number  of 
sides,  the  ratio  of  the  areas  is  equal  to  the  ratio  of  the  squares  of  the 
radii  or  of  the  apothems. 

MEASUREMENT   AND   EQUIVALENCE 
To  measure  a  segment  is  to  find  the  number  of  times  that  it  contains 

another  segment  which  is  taken  as  a  unit. 

The  number  found  is  called  the  measure  number,  the  measure,  or 

the  length  of  the  segment. 

Two  segments  are  said  to  be  commensurable  if  they  can  be  measured 

exactly  by  a  common  unit  of  measure. 


200  SOLID   GEOMETRY 

Two  segments  are  said  to  be  incommensurable  if  there  is  no  common 
unit  that  will  measure  each  exactly. 

An  irrational  number  is  a  number  that  cannot  be  expressed  as  an 
integer  or  as  the  quotient  of  two  integers. 

To  measure  the  surface  inclosed  by  the  sides  of  a  polygon  is  to  find 
how  many  times  it  contains  another  surface  chosen  as  a  unit  of  measure. 

The  area  of  a  polygon  is  the  measure  number  of  the  surface  of  the 
polygon. 

Two  polygons  that  cover  the  same  extent  of  surface  are  called 
equivalent  polygons.  The  symbol  (  =  )  is  used  for  equivalence. 

As.  59.  If  equivalent  polygons  are  added  to  equivalent  polygons, 
the  results  are  equivalent  polygons. 

As.  GO.  If  equivalent  polygons  are  subtracted  from  equivalent 
polygons,  the  results  are  equivalent  polygons. 

As.  61.  If  equivalent  polygons  are  divided  into  the  same  number 
of  equivalent  polygons,  each  part  of  one  is  equivalent  to  any  part  of 
the  other. 

As.  62.  Polygons  equivalent  to  the  same  polygon  or  to  equivalent 
polygons  are  equivalent. 

Any  two  polygons  are  equivalent  if  they  are  sums,  differences,  or 
equal  parts  of  equivalent  polygons. 

As.  63.  The  number  of  units  of  area  in  a  rectangle  is  equal  to  the 
product  of  the  number  of  units  of  length  in  the  base  and  altitude. 

THEOREM  113.  The  area  of  a  parallelogram  is  the  product  of  the 
base  and  altitude. 

THEOREM  114.  The  area  of  a  triangle  is  one-half  the  product  of 
the  base  and  altitude. 

THEOREM  115.  The  area  of  a  trapezoid  is  equal  to  one-half  the  prod- 
uct of  the  altitude  and  the  sum  of  the  bases. 

THEOREM  116.     Two  parallelograms  or  two  triangles  are  equivalent  if 

1.  They  have  equal  bases  and  are  between  the  same  parallels. 

2.  a=af  and  b  =  b'. 

3.  ab  =  a'bf. 

THEOREM  117.  If  a  triangle  and  a  parallelogram  have  equal  bases 
and  equal  altitudes,  the  triangle  is  equivalent  to  half  the  parallelogram- 

THEOREM  118.  The  square  constructed  on  the  hypotenuse  of  a 
right  triangle  is  equivalent  to  the  sum  of  the  squares  constructed  on 
the  other  two  sides. 

The  length  or  the  circumference  of  a  circle  is  defined  as  the  limit 
of  the  perimeters  of  a  series  of  polygons  inscribed  in  or  circumscribed 
about  a  circle  as  the  number  of  sides  is  increased  indefinitely. 


OUTLINE   OF   PLANE  GEOMETRY  201 

As.  64.  The  limit  of  the  perimeters  of  a  series  of  regular  polygons 
inscribed  in  or  circumscribed  about  the  same  circle  as  the  number  of 
sides  is  increased  indefinitely  is  the  same. 

This  limit  does  not  depend  upon  the  number  of  sides  of  the  initial 
polygon  nor  upon  the  method  of  increasing  the  number  of  sides.  This 
limit  is  TTd. 

THEOREM  137.    The  circumference  of  a  circle  of  diameter  d  is  wd. 

The  area  of  a  circle  is  defined  as  the  limit  of  the  areas  of  a  series  of 
inscribed  or  circumscribed  regular  polygons  as  the  number  of  sides  is 
increased  indefinitely. 

As.  65.  The  limit  of  the  areas  of  a  series  of  regular  polygons 
inscribed  in  or  circumscribed  about  the  same  circle  as  the  number  of 
sides  is  increased  indefinitely  is  the  same.  This  limit  is  one-half  the 
product  of  the  radius  of  the  circle  and  its  circumference. 

THEOREM  138.  The  area  of  a  circle  is  one-half  the  product  of  its 
radius  and  circumference. 

A  sector  of  a  circle  is  a  figure  bounded  by  two  radii  and  the  sub- 
tended arc. 

As.  66.  The  area  of  a  sector  has  the  same  ratio  to  the  area  of  a  circle 
of  which  it  is  a  part  as  the  angle  of  the  sector  has  to  four  right  angles. 

THEOREM  139.  The  ratio  of  the  circumference  to  the  diameter  is 
the  same  for  all  circles. 

THEOREM  140.  The  ratio  of  the  circumferences  of  two  circles  equals 
the  ratio  of  their  diameters  or  of  their  radii. 

THEOREM  141.  The  ratio  of  the  areas  of  two  circles  equals  the  ratio 
of  the  squares  of  their  radii  or  of  their  diameters. 


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Theoretically  and  Historically.     D.  C.  Heath  &  Co. 
Flatland.     A  Romance  of  Many  Dimensions.     Little,  -Brown  &  Co. 

202 


REFERENCES  AND   TOPICS  203 

GODFREY,  C.,  and  SIDDONS,  A.  W.     Modern  Geometry.    Putnam. 
HALL,  HENRY  SINCLAIR,   and   KNIGHT,  S.  R.      Higher  Algebra. 

Macmillan. 
HALSTED,  GEORGE  BRUCE.     Metrical  Geometry.     An  Elementary 

Treatise  an  Mensuration.     Ginn  &  Co. 

HANSTEIN,  HERMANN.     Constructive  Drawing*  Keuffel  &  Esser. 
Heath's  Mathematical  Monographs.     Issued  under  the  editorship  of 

Webster  Wells.     D.  C.  Heath  &  Co. 
HENRICI,  OLAUS  M.  F.  E.     Congruent  Figures.     Longmans,  Green 

&  Co. 
HILL,  GEORGE  FRANCIS.     The  Development  of  Arabic  Numerals  in 

Europe,   Exhibited  in   Sixty-four   Tables.     Clarendon  Press, 

Oxford,  England. 
HINTON,  CHARLES  HOWARD.    Scientific  Romances*    John  Lane 

&  Co. 
HINTON,    CHARLES    HOWARD.     The    Fourth    Dimension*     John 

Lane    &    Co. 

JOHNSON,  WILLIS  ERNEST.  Mathematical  Geography.  American 
Book  Co. 

JONES,  SAMUEL  ISAAC.  Mathematical  Wrinkles  for  Teachers  and 
Private  Learners*  The  Author,  Gunter,  Texas. 

KARPINSKI,  Louis  CHARLES.  Robert  of  Chester's  Latin  Transla- 
tion of  the  Algebra  of  Al-Khowarizmi  Macmillan. 

KLEIN,  FELIX.  Famous  Problems  of  Elementary  Geometry.  Ginn 
&  Co. 

LANGLEY,  EDWARD  MANN.  A  Treatise  on  Computation.  Long- 
mans, Green  &  Co. 

LIGHTFOOT,  JOHN.  Studies  in  Graphic  Arithmetic.  Normal  Cor- 
respondence College  Press,  London. 

LOCK,  JOHN  BASCOMBE.     Elementary  Statics*    Macmillan. 

MANNING,  HENRY  PARKER.  The  Fourth  Dimension  Simply 
Explained.  Munn. 

Mathematical  Gazette. 

MORRISON,  GABRIEL  JAMES.  Maps,  Their  Uses  and  Construction. 
Edward  Stanford,  London. 

*  Out  of  print. 


204  SOLID  GEOMETRY 

OLCOTT,  W.  T.     Star  Lore  of  All  Ages.     Putnam. 

PEKER,  CHARLES  GODFREY.      The  Steel  Square  as  a  Calculating 

Machine.     Industrial  Book  Co. 
PHILLIPS,  ANDREW  WHEELER,  and  FISHER,  IRVING.      Plane  and 

Solid  Geometry.    American  Book  Co. 
RIETZ,  HENRY  LEWIS,  and  CRATHORNE,  A.  R.     College  Algebra, 

Henry  Holt  &  Co. 
Row,  T.  SUNDARA.     Geometric  Exercises  in  Paper  Folding.     Open 

Court  Publishing  Co. 
School  Science  and  Mathematics. 
SCHUBERT,    HERMANN.      Mathematical    Essays    and    Recreations. 

Translated  from  the  German  by   Thomas  J.    McCormack. 

Open  Court  Publishing  Co. 

SCHULTZE,  ARTHUR.     Graphic  Algebra.    Macmillan. 
SMITH,  DAVID  EUGENE,  and  KARPINSKI,  Louis  CHARLES.     The 

Hindu- Arabic  Numerals.     Ginn  &  Co. 
STARK,  W.   E.   "Measuring  Instruments  of  Long  Ago,"  School 

Science  and  Mathematics,  February,  1910. 
SYKES,  MABEL.    A  Source  Book  of  Problems  for  Geometry.    Allyn 

&  Bacon. 

Teachers  College  Record. 
WENTWORTH,  GEORGE  ALBERT.     Plane  and  Solid  Geometry  (1889). 

Ginn  &  Co. 
WHITE,  U.   F.    Scrap  Book  of  Elementary  Mathematics.     Opea 

Court  Publishing  Co. 
WHITEHEAD,    ALFRED    NORTH.      Introduction    to    Mathematics. 

Henry    Holt    &    Co. 
WILSON,    THOMAS.      "The    Swastika,"    Smithsonian    Institution 

Annual    Report,    1894. 

Wonderland.     (Chinese    Monad.)     1901.     Great   Northern  Rail- 
road. 
YOUNG,    JACOB    WILLIAM    ALBERT.      Monographs    on    Modern 

Mathematics.     Longmans,    Green    &    Co. 
YOUNG,   JOHN  WESLEY.    Lectures  on   Fundamental   Concepts  of 

Algebra   and  Geometry.     Macmillan. 


REFERENCES  AND   TOPICS  205 

TOPICS 

HISTORY 

Bolshevik  Multiplication.     School  Science  and   Mathematics,    Vol. 

XVIII,  p.  698,  Vol.  XIX,  p.  359. 
Development  of  Algebraic  Symbolism.  Any  history  of  mathematics. 

See  J.  W.  Young,  p.  226. 
Finger  Reckoning.     Magazine  articles.    See  American  Mathematical 

Monthly,    "Digital    Reckoning    among   the   Ancients,"  Vol. 

XXIII,  p.  7. 

Hindu- Arabic  Numerals.     Smith  and  Karpinski;  Hill. 

History  of  Decimals.     Cajori,  p.  150. 

Methods  of  Multiplying,  Historic  and  Primitive.      Cajori. 

Napier's  Rods.  Ball,  .1  Short  Account  of  the  History  of  Mathe- 
matics; White,  p.  69. 

Number  Systems  of  the  North  American  Indians.  American 
Mathematical  Monthly,  November,  1913.  pp.  263,  293. 

Rhetorical  Algebra.  Karpinski.  This  includes  geometric  solutions 
of  quadratics. 

The  Abacus  and  Its  Uses:  The  Roman  Abacus.  Any  history  of 
elementary  mathematics. 

ARITHMETIC 

Approximate  Computations.     Evans  and  Marsh,  p.  50;  Langley. 
Duodecimals.     School  Science  and  Mathematics,  Vol.  IX,  pp.  516, 

555. 
Multiplication  Curiosities.     Jones,  pp.  71,  73.  77;    Teachers  College 

Record;  White,  p.  17. 
Nine  and  Its  Properties;    Casting  Out  Nines.     Bemen  and  Smith, 

p.  19;   Jones,  p.  63,  Ex.  33;   p.  81,  Ex.  159. 
Prime  Numbers.     Jones,  p.  86. 
Repeating  Decimals.     Educational  Bi-Monthly,  Vol.   IX,  p.  408; 

White,  p.   11. 
Short   Methods    of   Computation.      Teachers   College   Record,   Vol. 

XIII,   p.   385. 

Systems  of  Notation.     Hall  and  Knight. 
14 


206  SOLID    GEOMETRY 

Tests  for  Divisibility.     Bemen  and  Smith. 

Tricks.  Jones,  p.  63,  Ex.  34;  p.  67,  Ex.  61;  p.  81,  Ex.  157;  p.  84, 
Ex.  177;  p.  103,  Ex.  249;  White,  p.  19. 

ALGEBRA 

Engineering  Problems  Giving  Rise  to  Equations,  Quadratic  and 
Higher.  Rietz  and  Crathorne.  pp.  68,  140. 

Fallacies.    Jones;   White. 

Graphic  Solution  of  Equations.     Schultze. 

Graphic  Solution  of  Problems.     Lightfoot. 

Imaginaries.     Whitehead;   J.  W.  Young. 

Interest  and  Annuities. 

Tricks.  Ball,  Mathematical  Recreations  and  Essays,  pp.  4-13; 
Jones,  p.  58,  Ex.  5,  6;  p.  62,  Ex.  30;  p.  63,  Ex.  31,  32;  p.  67, 
Ex.62;  p.  76,  Ex.  131,  133j  p.  79,  Ex.  147;  p.  81,  Ex.  158,  161; 
p.  87,  Ex.  190;  p.  89,  Ex.  200,  201;  White,  p.  25. 

GEOMETRY 

Construction  of  Regular  Polygons  by  Paper  Folding.     Row. 
Geometric  Fallacies.     Ball,  Mathematical  Recreations  and  Essays; 

Jones. 
Proofs  for  Pythagorean  Theorem.     American  Mathematical  Monthly, 

1896-99. 

Pythagorean  Numbers.     Encyclopedia  Britannica;  Halsted. 
Star  Polyhedra.     Mathematical  Gazette,  Vol.  VIII,  p.  269. 
The  Triangle  and  Its  Circles.     Bruce. 

HIGHER   MATHEMATICS 

Coaxial  Circles.     Godfrey  and  Siddons,  pp.  87-96. 

Conic   Sections,   Construction   and   Properties   of.      Row,   p.    116; 

School  Science  and  Mathematics,  Vol.  VII,  p.  595;  Wentworth 

(1889),  chap.  ix. 

Fourth  Dimension.     Flatland;   Hinton;   Manning. 
Higher   Curves,   Construction  and   Uses  of.     American  School   of 

Correspondence  Circular,  Mechanical  Drawing,  II;    Becker; 

Hanstein;   Row;   Scientific  American,  Sept.  23,  1916. 
Nine-Point  Circle.     Godfrey  and  Siddons. 


REFERENCES  AND   TOPICS  207 

Permutations  and  Combinations.     Jones,  pp.  104,  105. 

Sum  of   the  Angles   of  a    Triangle   on  Sphere  and  Pseud osph ere. 

J.  W.  A.  Young,  p.    110. 

The  Pseudosphere.     Phillips  and  Fisher,  p.  526. 
Trisection  of  Any  Angle.     Row,  pp.  135,  143;   School  Science  and 

Mathematics,  Vol.  X,  p.  582;    Vol.  XIII,  p.  546. 

MISCELLANEOUS 

Ciphers   and   Cryptography.     Ball,    Mathematical    Recreations   and 

Essays. 
Geometry  and  Arabic  Art.     Sykes,  Source  Book  for   Problems  of 

Geometry.     See  Index. 
Gothic  Windows.     Sykes,  Source  Book. 
Graphs  and  Railroad  Time  Tables. 
History  of  the  Calendar.     Johnson,  chap.  vii. 
Magic  Squares.     Andrews. 
Map  Making.     Adams;   Johnson;   Morrison. 
Measuring  Instruments,  Ancient  and  Modern.     Stark. 
Paradromic  Rings.     Ball,  Mathematical  Recreations  and  Essays. 
Primitive    Geometric    Figures;    Pentagram    Star,   Chinese    Monad, 

Swastika. 

Rectification  of  the  Circle.     Sykes-Comstock   Plane  Geometry,  p. 

282. 

The  Steel  Square.     Fair. 
Tiles,  Origin  and  Shapes.     Sykes,  Source  Book. 


THE  INDEX 


[References  are  to  page  numbers] 


Addition,  or  composition..  .  .    195 

Adjacent  dihedral  angles ....     23 

Alternation:   extreme 195 

mean 195 

Altitude,  length  of:    in  equi- 
lateral triangle 197 

of  cone 64 

of  cylinder 55 

of  frustum  of  pyramid ....     63 

of  parallelogram 189 

of  prism 51 

of  prismatoid '.   121 

of  pyramid 60 

of  triangle 195 

of  zone 145 

Angle:     central,    of    regular 

polygons 199 

conical 38 

corresponding  central  poly- 
hedral       89 

dihedral 21 

of  line  and  plane 37 

of  sixty  degrees 187 

plane,  of  dihedral 22 

polyhedral 38 

solid 38 

spherical 86 

trihedral 38 

Angles:  adjacent  dihedral. ..     23 

and  parallels 187 

complementary  dihedral .  .      23 

congruent  dihedral 22 

congruent  polyhedral 38 

equal  dihedral 22 

equal  in  space 17 

face,  of  polyhedral  angle .  .     38 

of  spherical  polygon 89 

supplementary  dihedral ...     23 

symmetric  polyhedral 39 

vertical  dihedral 23 

vertical  trihedral 97 

Angle-sums 187 

in  spherical  triangles 94 

Apothem  of  regular  polygon .    199 


Approximation    to   area:     of 

cone 130 

of  cylinder 127 

of  frustum  of  cone 132 

of  sphere 142 

Approximation    to    volume: 

of  cone 131 

of  cylinder 128 

of  frustum  of  cone 132 

of     rectangular     parallele- 
piped    Ill 

of  sphere 150 

Area:   of  circle 201 

of  lune 146-147 

of  plane  figures 200 

of  polygons 200 

of  sphere 142-144 

of  spherical  triangle .  .  .  148-149 

of  zone 145-146 

units  of 183 

(See  also  Lateral  areajTotal 
area) 

Areas:  formulae  for 156 

ratio  of 168,  170,  171 

in  plane  figures 197 

Assumptions         concerning: 

cones 66 

cylinders 57,  126 

dihedral  angles 22 

equivalent  spherical  poly- 
gons     101 

equivalent  spherical  pyra- 
mids     119 

lines 4,  15 

lunes 147 

planes 2,  3,  15 

prisms  inscribed   in  cylin- 
ders. .......... 126 

prisms  inscribed  in  pyra- 
mids    119 

similar  solids 166 

spheres 76,  77 

spherical  pyramids 154 

spherical  triangles 96,  97 


209 


210 


THE   INDEX 


Assumptions  concerning : 

spherical  wedges 154 

trihedral  angles 96,  97 

volumes 109,  119 

Axial  symmetry 173,  188 

Axis:  of  circle  of  sphere 82 

of  cone  of  revolution 67 

of  cylinder 74 

of  cylinder  of  revolution .  .  59 

of  frustum  of  revolution .  .  69 

of  pencil  of  planes 3 

of  symmetry 173 

of  symmetry  in  plane  geom- 
etry  ! 188 

Base :  of  cone 64 

of  pyramid 60 

Bases:   of  cylinder 55 

of  prism 51 

of  zone 145 

Birectangular    spherical    tri- 
angle    95 

Blank  (sheet-metal  work) .  .  16p 

Capacity,  units  of 184 

Cavalieri's  theorem 138 

volume  by 138-140,  153 

Center:    of  gravity  of  plane 

triangle 195 

of  parallelepiped 53 

of  regular  polygon 199 

of  similitude 164,  197 

of  sphere 76 

of  spherical  surface 76 

radial 163 

Central  angle  of  regular  poly- 
gon   199 

Central  polyhedral  angle ....  89 

Central  projection 35 

Central  symmetry 188 

Circle:  area  of 201 

circumference,  of 200 

inscribed. 198 

Circle  of  sphere 81 

axis  of 82 

great 81 

small 81 

polar  distance  of 83 

poles  of 82 

Circles:   and  related  angles.  .  192 

and  related  lines 192 

tw.o .194 


Circular  cone 67 

volume  of 131,  140,  156 

Circular  cylinder 58 

lateral  area  of 126-127,  156 

Circumcenter    of    plane    tri- 
angle    195 

Circumference  of  circle 200 

Circumscribed  circle 198 

Circumscribed     cone     about 

pyramid 129 

Circumscribed  conical  surface 

about  sphere 104 

Circumscribed  cylinder  about 

prism 125 

Circumscribed  cylindrical  sur- 
face about  sphere-. 104 

Circumscribed  polygon 1 98 

Circumscribed  polyhedron ...     79 

Circumscribed  sphere 79 

Closed  surface 50 

Commensurable  segments ...    199 

Complementary  angles 185 

Complementary          dihedral 

angles 23 

Composition,  or  addition ....    195 
Concave  polyhedral  angle.  .  .     38 

Cone 64 

altitude  .of 64 

approximate  lateral  area  of  130 
approximate  volume  of ...    130 

base  of 64 

circular  (See  Circular  cone) 

circumscribed 104,  129 

convex 66 

elements  of 64 

frustum  of  (See  Frustum  of 
cone) 

lateral  area  of 130,  134,  156 

lateral  surface  of 64 

oblique  circular 67 

of  revolution 67 

properties  of 64 

pyramid  inscribed  in 129 

right   circular    (See    Right 
circular  cone) 

sections  of 04,  66 

spherical 154 

total  area  of '.  .  .    130 

truncated 68 

vertex  of • 64 

volume  of 131,  IK),  ir.li 

with   circular   base 

131,  140,  156 


THE   INDEX 


211 


Cone  of  revolution 67 

axis  of 67 

lateral  area  of 134,  156 

volume  of 131,  140,  156 

Cones,    similar    (See   Similar 

cones  of  revolution) 

Congruent  dihedral  angles ...  22 
Congruent  polyhedral  angles 

38,  41 

Congruent  right  prisms 113 

Congruent  solids 112 

Congruent  triangles :   plane..  186 

spherical 96-101 

Congruent  trihedral  angles 

41,96-101 

Congruent  truncated  prisms.  112 

Conical  angle 38 

Conical  space 64 

Conical  surface 49 

circumscribed  about  sphere  104 

nappes  of 50 

of  revolution 34 

tangent  to  sphere 104 

vertex  of 49 

Construction :      of    inscribed 

frustums  of  pyramids .  .  .  132 

of  inscribed  polyhedrons. . .  142 

of  inscribed  prisms.  .  .  .118,  125 

of  inscribed  pyramids 129 

of  perpendicular  lines  and 

planes 44 

of     rectangular     parallele- 
piped    54 

of  regular  polygons 198 

of  regular  polyhedrons.  ...  71 

of  star  polyhedrons 72 

Contact,  point  of 77 

Convex  cone 66 

Convex  cylinder 55 

Convex  polyhedral  angle ....  38 
Convex  spherical  polygon .  .  89,  90 

Convex  surface 50 

Corresponding   central    poly- 
hedral angle '.  .  89 

Cosines,  table  of 182 

Cube 55 

Cube  root 181 

Cube  roots,  table  of 181 

Curved  surface 50 

Cylinder 55,  58 

altitude  of 55 

approximate  lateral  area  of.  127 

approximate  volume  of ...  128 


axis  of 74 

bases  of 55 

circular  (See  Circular  cyl- 
inder) 

circumscribed :.  104,  125 

convex 55 

elements  of 55 

elliptical 58 

lateral  area  of.  . .  .127,  134,  156 

lateral  surface  of 55 

naming  of 58 

oblique 59 

of  revolution  (See  Cylinder 
of  revolution) 

prisms  inscribed  in 125 

properties  of 56 

right... ..     59 

right    circular    (See    Right 
circular  cylinder) 

sections  of 56-58 

total  area  of 127,  156 

truncated 55 

volume  of 128,  140,  156 

with    circular  base 

128,  140,  156 

Cylinder  of  revolution 59 

axis  of 59 

lateral  area  of 134,  156 

volume  of 128,  140,  156 

Cylinders,  similar  (See  Sim- 
ilar cylinders  of  revolu- 
tion) 

Cylindrical  space 55 

section  of 56 

Cylindrical  surface . 49 

circumscribed  about  sphere  104 

of  revolution 32 

right  section  of 51 

tangent  to  sphere 104 

Decagon 198 

Definite  location  (See  Deter- 
mination) 

Degree,  spherical 147 

Determination:  of  lines 

12,  15,  26,  42 

of  planes 3,  14,  28,  42 

of  spheres 79 

Development  of  lateral  sur- 
face: of  frustum 135 

of  right  circular  cone 130 

of  right  circular  cylinder .  .    127 
Diameter  of  sphere .  .  76,  103,  Ex. 


212 


THE    INDEX 


Dihedral   angle 21 

edges  of 21 

faces  of 21 

measure  of 22 

plane  angle  of 22 

Dihedral  angles:  adjacent. . .  23 

complementary 23 

congruent 22 

equal.... v..  22 

of  polyhedral  angle 38 

related 23 

supplementary 23 

vertical 23 

Direct  radial  position 165 

Directly  similar,  definition. . .  165 

Director  of  surface 50 

Distance:   from  point  to  plane  16 

polar 83 

spherical 82 

Division  or  subtraction 196 

Dodecahedron,  regular 69 

Duodecagon 198 

Edge:  of  dihedral  angle 21 

of  half-plane 21 

Edges:  lateral,  of  prism 51 

lateral,  of  pyramid 60 

of  polyhedral  angle 38 

of  polyhedrons 48 

of  prismatic  surface 50 

of  pyramidal  surface 50 

Element  of  a  surface 50 

Elements :   of  cone 64 

of  cylinder 55 

Elliptical  cylinder 58 

Equal  angles  in  space 17 

Equal  dihedral  angles 22,  163 

Equal  ratios,  tests  for 196 

Equal  segments  from  point  to 

plane 13 

Equations 178 

Equilateral  spherical  triangle.  89 
Equivalent   figures   of   plane 

geometry 199 

Equivalent  prisms 113 

Equivalent  pyramids 119 

Equivalent  sections 61,  65 

Equivalent  solids 112 

Equivalent  triangular  prisms.  115 
Equivalent   triangular   pyra- 
mids  119,  139 

Excess,  spherical 95 

Extreme  alternation .  .  .195 


Face    angles    of    polyhedral 

angle 38 

Faces:   of  dihedral  angle ....     21 

of  polyhedral  angle 38 

of  polyhedron 48 

Formulae  for  areas  and  vol- 
umes     156 

Fractions 174 

Frustum  of  cone 68 

properties  of 69 

Frustum  of  cone  with  circular 

base,  volume  of 132 

Frustum  of  revolution 69 

axis  of 69 

Frustum    of    right    circular 
cone:       development    of 

surface  of 135 

lateral  area  of 132,  135,  156 

properties  of 69 

slant  height  of 69 

volume  of 132,  156 

Frustum  of  pyramid 63 

altitude  of 63 

properties  of 63 

volume  of 123,  156 

Frustum  of  regular  pyramid: 

lateral  area  of 108,  156 

properties  of 63 

slant  height  of 63 

Frustum  of  revolution 69 

axis  of 69 

lateral  area  of .  .  .  .132,  135,  156 

volume  of 132,  156 

Fundamental         assumption 

concerning  volumes 109 

Fundamental     characteristic 

of  planes 2. 

Fundamental  tests:    for  par- 
allel lines  and  planes .... 
for  parallel  planes.  .  .  . 
for  perpendicular  lines  and 

planes 12 

for  perpendicular  planes. .  .     23 

General  spherical  polygons.  . 

Generator  of  surface 50 

Geometry,     spherical,     com- 
pared with  plane  87,  89,  105 
Gravity,  center  of,  of  triangle.  195 
Great  circle  of  sphere 81 


Half-plane 


21 


THE   INDEX 


213 


Height,     slant     (See     Slant 
height) 

Heptagon 198 

Hexagon 198 

Hexahedron,  regular 69 

Icosahedron,  regular 69 

Incenter  of  plane  triangle 195 

Inclination  of  line  and  plane.  37 
Incommensurable,  definition.  200 
Inequalities  in   plane  geom- 
etry    190 

Inscribed  circle : 198 

Inscribed  polygons 198 

Inscribed  polyhedrons 79 

Inscribed  prisms  in  cylinders  125 

Inscribed  prisms  in  pyramids  118 

Inscribed  pyramids  in  cones. .  129 

Inscribed  spheres 79 

Intersection  of  two  planes. . .  .3,  4 

Irrational  numbers 200 

Isosceles  spherical  triangle  89,  100 

Isosceles  trapezoid 189 

Isosceles  triangle 188 

Inverse  radial  position 165 

Inversely  similar,  definition. .  165 

Inversion .  .                               .  195 


Lateral  area,  formulae  for:       156 

of  circular  cylinder 126,  127 

of  cone  of  revolution 134 

of  cylinder  of  revolution. . .    134 
of  frustum  of  regular  pyra- 
mid    108 

of  frustum  of  right  circu- 
lar cone 132,  135 

of  polyhedron 106 

of  prism 106 

of  regular  pyramid 106 

of  right  circular  cone 

129,  130,  134 

of  right  circular  cylinder .  . 

127,  134 

of  solids  of  revolution 134 

Lateral  edges:   of  prism 51 

of  pyramid 60 

Lateral  surface :  of  cone ....     64 

of  cylinder 55 

of  prism 51 

of  pyramid 60 

Lateral     surfaces,     develop- 
ment of 127,  130,  135 


Length :  of  segment 199 

units  of 183 

Line 2 

angle  with  plane 37 

as  locus 32 

inclination  to  plane 37 

lying  in  plane,  tests  for 

2,  8,  15,  26,  42 

projection  of,  on  plane. ...     35 
tangent  to  sphere 77 

Lines:   determination  of 

15,  16,  26,  42 

on  spheres 87-88 

parallel 4,  11,  43 

parallel  to  planes 4,  11,  43 

perpendicular 13,  43 

perpendicular  to  planes 

12,  13,  19,  25,  27,  43 

ratios  of 167 

related  to  circles 192 

relations  to  planes 4 

relative  positions  of 4 

skew 4 

Location,  definite  (See  Deter- 
mination) 

Locus :  in  plane  geometry ...    1 94 
in  space 29 

Lune 89 

area  of 146,  147 

ratio  of  surface  to  surface 
of  sphere 147 


Mean  alternation 195 

Mean  proportional 196 

Measure:   of  segments 199 

of  surface 109,  200 

Measurement:  of  angles.  ...  193 

of  cones 129 

of  cylinders 125 

of  dihedral  angles 22 

of  frustums  of  cones 132 

of  polyhedral  angles 38 

of  round  bodies 125 

of  space 109 

of  spherical  angles 86 

(See  Areas;   Volumes) 
Measurements :       in       plane 

geometry 199 

spherical 142 

Median 189,  190 

Metric  units 184 

Mid-section  of  prismatoid.  .  .  121 


214 


THE   INDEX 


Naming :  of  cylinders 58 

of  prisms .  51 

Nappes :  of  conical  surface .  .  50 

of  pyramidal  surface 50 

Numbers,  irrational 200 

Oblique  circular  cone 67 

Oblique  cylinder 59 

Oblique  prism 53 

Octagon 198 

Octahedron,  regular 69 

Orthocenter  of  triangle ]  95 

Orthogonal  projection 35 

Parallel  lines,  tests  for 

4,  7,  8,  10,  11,  18,  19,  36,  43 
Parallel  lines  and  planes,  tests 

for 4,  6, 11,  43 

Parallel  planes,  tests  for 

3,  9,  11,20,42 

Parallel  projection 35 

Parallelepiped 53 

center  of 53 

properties  of 53 

rectangular  (See  Rectangu- 
lar parallelepiped) 

right 54 

volume  of 112 

Parallelogram 189 

Parallels:  and  angles 187 

and  transversals 190 

Pencil  of  planes 3 

axis  of , 3 

Pentadecagon 198 

Pentagon 198 

Perpendicular  lines 13,  43 

Perpendicular  lines  and  planes, 

tests  for.  12,13,18,19,25,27,43 
Perpendicular  planes,  tests 

for 23,24,43 

Plane 49 

angle  of  line  and 37 

containing  line  ..2,8,  15,  26,  42 
distance  from  point  to.  ...      16 

half. 21 

inclination  of  line  to 37 

projection  of  line  on 35 

projection  of  point  on 35 

tangent  to  sphere 77 

Plane  angle  of  dihedral  angle     22 
Plane     geometry,     spherical 

compared  with.  .  .87,  89.  105 
Plane  surface . .  48 


Planes:   determination  of 

3,  14,  15,28,42 
fundamental  characteristic 

of 2 

intersection  of 3,  4 

parallel 3,9,  11,  20,42 

parallel  to  lines 4,  6,  11,  43 

pencil  of '.       3 

perpendicular 23,  24,  43 

perpendicular  to  lines 

12,  13,  18,  19,  25,  27,  43 

relations  of,  to  lines 4 

relative  positions  of 3 

Point 2 

distance  from,  to  plane ....      16 

of  contact 77 

of  tangency 77 

projection  of 35 

Polar  distance 83 

Polar  triangles 91 

relations  of 92,  93 

Poles  of  circle  of  sphere 82 

Polygons  (plane):   area  of . .  .   200 

circumscribed 198 

inscribed 198 

radially  placed 197 

Polygons,  regular  (See  Regu- 
lar polygons) 

Polygons,  spherical  (See  Spher- 
ical polygons) 

Polyhedral  angle 38 

concave 38 

convex 38 

corresponding  central 89 

dihedral  angles  of 38 

edges  of 38 

face  angles  of 38 

faces  of 38 

measure  of 38 

properties  of 39,  40 

vertex  of 38 

Polyhedral  angles:       congru- 
ent      38 

symmetric 39 

Polyhedron 48 

area  of... 106 

circumscribed 79 

edges  of 48 

faces  of 48 

inscribed 79 

vertices  of 48 

Polyhedrons:      radial    center 

of..  .   163 


THE   INDEX 


215 


radially  placed 163 

regular  (See  Regular  poly- 
hedrons) 

similar   (See  Similar  poly- 
hedrons) 

star 72 

Position,  radial 165 

Positions:   of   planes 3 

of  straight  lines 4 

relative,  of  line  and  plane .  4 

Prism 51 

altitude  of 51 

bases  of 51 

inscribed  in  cylinder 125 

inscribed  in  pyramid 118 

lateral  area  of 106 

lateral  edges  of . . 51 

lateral  surface  of . 51 

naming  of .--. 51 

oblique 53 

properties  of 52 

regular 53 

right 53 

square 51 

truncated 53,  161 

volume  of.  .  .117,  123,  138,  156 

Prismatic  space 51 

section  of 52 

Prismatic  surface 49 

edges  of 50 

right  section  of 51 

Prismatoid 121 

altitude  of 121 

mid-section  of 121 

volume  of 122,  156 

Prisms:   equivalent 115 

inscribed  in  cylinders 125 

series  of  inscribed,  in  pyra- 
mids   118 

special 53 

Projection:   central 35 

of  line  on  plane 35 

of  point  on  plane 35 

of  segment 36 

orthogonal 35 

parallel 35 

Proportion 195,  196 

Proportional  segments .....  29,  60 

Pyramid 60 

altitude  of 60 

base  of 60 

frustum  of  (See  Frustum  of 
pyramid) 


inscribed  in  cone 129 

lateral  area  of 107,  156 

lateral  edges  of 60 

lateral  surface  of 60 

prisms  inscribed  in 118 

properties  of 60 

regular  (See  Regular  pyra- 
mid) 

section  of 60 

spherical      (See     Spherical 
pyramid) 

truncated 63 

vertex  of 60 

volume  of 121,  156 

Pyramidal  space 60 

Pyramidal  surface 49 

edges  of 50 

nappes  of 50 

vertex  of 49 

Pyramids,  equivalent 119,  139 

Quadrilaterals 189 

Radial  center 163 

Radially  placed  polygons ....    197 
Radially  placed  polyhedrons .   163 

Radial  position 163 

direct 165 

inverse 165 

Radius:  of  regular  polygon ..  199 

of  sphere 76 

of  spherical  surface 76 

Ratio  of  similitude 164 

Ratios:  general 195,  196 

methods  for  proving  equal.  196 
of  areas 

61,  65,  147,  168,  170,  171 

of  lines 167 

of  volumes...  154,  168,  169,  171 

Ray 192 

Reciprocal  relations 91 

Rectangle 189 

Rectangular  parallelepiped .  .     54 

volume  of 109 

Regular  polygons  (plane)..  .  .    198 

apothem  of 199 

center  of 199 

construction  of 198 

properties  of 198-199 

Regular  polyhedrons 69 

construction  of 71,  75,  Ex. 

number  of 70 

Regular  prism 53 


216 


THE   INDEX 


Regular  pyramid 62 

lateral  area  of 107 

properties  of 62 

slant  height  of 62 

Related  dihedral  angles 23 

Relative  positions:  of  lines..       4 

of  lines  and  planes 4 

of  planes 3 

Revolution:   cone  of.. 67,  134,  156 

conical  surface  of 34 

cylinder  of 59,  134,  156 

cylindrical  surface  of 32 

frustum  of 69 

sectors  of . . . , 152 

similar  solids  of 169 

solids  of 134,  136 

Rhombus 189 

Right  circular  cone 67 

approximate  lateral  area  of  130 
development  of  surface  of .    130 

lateral  area  of 130,  134,  156 

properties  of 67 

sections  of 65 

slant  height  of 67 

total  area  of 130,  156 

volume  of 131,  140,  156 

Right  circular  cylinder 59 

approximate  lateral  area  of  127 
development  of  lateral  sur- 
face of 127 

lateral  area  of.  .  .127,  134,  156 

properties  of 59 

total  area  of 127,  156 

volume  of 128,  140,  156 

Right  cylinder 59 

Right  parallelepiped 54 

Right  prism 53 

Right  section 51 

Roots  (See  Cube  root;  Square 
root) 

Section:  of  cone 64-66,  68 

of  cylinder 57,  58,  74,  Ex. 

of  cylindrical  space 56 

of  prismatic  space 52 

of  prisms; 73,  Ex. 

of  pyramid 60 

of  solid 48 

of  spherical  surface 81 

•right 51 

transverse 51 

Sections:  equivalent 61,  65 

of  right  circular  cone 68 


Sector:  of  circle 201 

spherical    (See    spherical 
sector) 

Segment:  measure  of 199 

projection  of 36 

spherical 155 

Segments,  proportional.  .  .  .29,  60 

Sides  of  spherical  polygbn ...     88 

Similar  cones  of  revolution .  .    169 

ratio  of  lateral  areas  of. ...    171 

ratio  of  total  areas  of 171 

ratio  of  volumes  of 171 

Similar  cylinders  of  revolu- 
tion      169 

ratio  of  lateral  areas  of 170 

ratio  of  total  areas  of 170 

ratio  of  volumes  of 171 

Similar  plane  figures 197 

center  of  similitude  of ....    197 

Similar  polyhedrons 162 

center  of  similitude  of 164 

directly  similar 165 

inversely  similar 165 

properties  of 166 

radial  center  of 163 

radially  placed 163 

ratio  of  areas  of 168 

ratio  of  similitude  of 164 

ratio  of  volumes  of 168,  169 

tests  for 162-165 

Similar  solids 164,  165 

properties  of 172 

Similar  tetrahedrons 162 

Sines,  table  of 182 

Skew  lines 4 

Slant  height:    of  frustum  of 

cone 69 

of  frustum  of  pyramid ....     63 

of  regular  pyramid 62 

of  right  circular  cone 67 

Small   circle   of   sphere    (See 
Circle  of  sphere) 

Solid 1,  48 

section  of 48 

Solid  angle 38 

Solids,  equivalent 112 

Solids  of  revolution:    lateral 

area  of 134 

similar  (See  Similar  solids) 

volume  of 136 

Space:   conical 64 

cylindrical 55 

measurement  of . .  .109 


THE   INDEX 


217 


prismatic 51 

Syramidal 60 
ere 76 

*  approximate  area  of 142 

approximate  volume  of. ...    150 

area  of 142,  144,  157 

center  of 76 

circumscribed 79 

determination  of 79 

diameter  of 76 

inscribed 79 

lines  on 87 

radius  of 76 

tangents  to 77,  104 

volume  of.  .  .150,  151,  153,  157 
Spheres:  determination  of ...     79 

ratio  of  areas  of 171 

ratio  of  volumes  of 171 

Spherical  angle 86 

measure  of 86 

sides  of 86 

vertex  of 86 

Spherical  cone 154 

Spherical  degree 147 

Spherical  distance 82 

Spherical  excess 95 

Spherical  geometry  compared 

with  plane 87,  89,  105 

Spherical  measurements ......   142 

Spherical  polygons 88 

angles  of 89 

convex 89 

corresponding  central  poly- 
hedral angle  of 89 

general 88 

sides  of 88 

vertices  of _ 88 

Spherical  pyramid 154 

volume  of 154,  157 

Spherical  sector 152 

volume  of 152,  157 

Spherical  segment 155 

volume  of 155,  157 

Spherical  surface 32,  76 

center  of 76 

of  revolution 32 

radius  of 76 

Spherical  triangle 89 

area  of 148,  149 

birectangular 95 

central  trihedral  angle  of . .     89 

equilateral 89 

isosceles 89,  100 


polar    of    (See    Polar    tri- 
angles) 

properties  of 90 

spherical  excess  of 95 

sum  of  angles  of 94 

trirectangular 95 

Spherical  triangles :  congruent  96 

equivalent 101,  102 

polar 91 

symmetric 96 

vertical 97 

Spherical  wedge 154 

volume  of 154 

Square 189 

Square  prism 51 

Square  root 175 

Square  roots,  table  of 180 

Star  polyhedrons 72 

Subtraction  or  division 196 

Supplementary  angles 185 

Supplementary  dihedral 

angles 23 

Surface 1,  48 

as  locus 32 

closed 50 

conical  (See  Conical  surface) 

convex 50 

curved 50 

cylindrical  (See  Cylindrical 
surface) 

director  of 50 

element  of 50 

generator  of 50 

lateral  (See  Lateral  surface) 

measurement  of 200 

plane 48 

prismatic     (See    Prismatic 

surface) 
pyramidal  (See  Pyramidal 

surface) 

spherical  (See  Spherical  sur- 
face) 

units  of 183 

Surface  of  revolution:    area 

of 134 

conical 34 

cylindrical 32 

spherical 32 

zone 145 

Symmetric  polyhedral  angles.  41 

Symmetric  polyhedrons 172 

Symmetric      spherical      tri- 
angles  96-100 


218 


THE   INDEX 


Symmetric    trihedral    angles 

41,  96-100 

Symmetry 172,  188 

Tables:  of  cube  roots 181 

of  sines,  cosines,  and  tan- 
gents   182 

of  square  roots 180 

of  units  of  capacity 184 

of  units  of  length 183 

of  units  of  surface 183 

of  units  of  volume 183 

of  units  of  weight 184 

Tangency,  point  of 192 

Tangent  to  circle 192 

Tangent  to  sphere:       conical 

surface 104 

cylindrical  surface 104 

lines 77 

planes 77 

tests  for 78 

Tangents,  table  of 182 

Tests  (See  various  congruent, 
equal,  equivalent,  paral- 
lel, perpendicular,  pro- 
portional, symmetric, 
and  tangent  figures) 

Tetrahedron,  regular 69 

Tetrahedrons,  similar. ......  162 

Total   area:     of    cone,    right 

circular 130,  156 

of  cylinder,  right  circular  127, 156 

Transverse  section 51 

Trapezoid 189 

isosceles 189 

Triangle  (See  Spherical  tri- 
angles) 

Triangular  prism,  volume  of. .  1 16 
Triangular  pyramid,  volume 

of 120 

Trihedral,  angles 38 

congruent 41,  96-100 

symmetric 41,  96-100 

vertical 97 

Trirectangular  spherical  tri- 
angle    95 

Truncated  cone 68 

Truncated  cylinder 55 

Truncated  prism 53,  161,  Ex. 

Truncated  pyramid 63 


Units:    metric 

of  area... 


.183-184 
.   183 


of  capacity 184 

of  length 183 

of  volume 183 

of  weight 184 

Vertical  dihedral  angles 23 

Vertical  spherical  triangles .  .      97 
Vertical  trihedral  angles ....      97 

Vertices:  of  polyhedron 48 

of  spherical  polygon 88 

Vertex:  of  cone 64 

of  polyhedral  angle 38 

of  pyramid GO 

Volume:  formulae  for,  listed.  156 

general  formula  for 121 

of  cone  with  circular  base 

131,  140 
of    cylinder    with    circular 

base .....128,  140 

of  frustum  of  cone 132 

of  frustum  of  pyramid ....    123 

of  parallelepiped 112 

of  prism 117-123,  138 

of  prismatoid 122 

of  pyramid 121-123 

of  rectangular  parallele- 
piped     109 

of  spheres 150,- 151,  153 

of  spherical  cone 154 

of  spherical  pyramid 154 

of  spherical  sector 152 

of  spherical  segment 155 

of  spherical  wedge 154 

of  triangular  prism 115 

of  triangular  pyramid 120 

of     truncated      triangular 

prism 161 

units  of 183 

Volumes:     by    Cavalieri's 

theorem 138,  153 

fundamental      assumption 

concerning 109 

ratios  of 168,  169,  171 

Wedge 157,  Ex. 

spherical • 154 

Weight,  units  of 184 

Zone 145 

altitude  of 145 

area  of 145,  146 

bases  of 145 

of  one  base 145 


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